Chapter7 resolve geometry Module2 straight line lower

Jump to heading 9.Positional Relationships Between Two Lines

Positional Relationship	Slope-Intercept Form $l_{1} : y = k_{1} x + b_{1}$ $l_{2} : y = k_{2} x + b_{2}$	General Form $l_{1} : a_{1} x + b_{1} y + c_{1} = 0$ $l_{2} : a_{2} x + b_{2} y + c_{2} = 0$
Parallel	$l_{1} ∥ l_{2} \Leftrightarrow k_{1} = k_{2}, b_{1} \neq b_{2}$	$l_{1} ∥ l_{2} \Leftrightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Intersecting	$k_{1} \neq k_{2}$	$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
Perpendicular	$l_{1} ⊥ l_{2} \Leftrightarrow k_{1} k_{2} = - 1$	$l_{1} ⊥ l_{2} \Leftrightarrow \frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} = - 1 \Leftrightarrow a_{1} a_{2} + b_{1} b_{2} = 0$

Examples

Slope-Intercept Form
- Parallel
  $\begin{array}{ll} y = 2 x - 1 \\ y = 2 x + 3 \\ k \to 2 = 2 \\ b \to - 1 \neq 3 & Not coincident \end{array}$
- Perpendicular
  $\begin{array}{ll} y = - \frac{1}{2} x - 1 \\ y = 2 x + 3 \\ k_{1} k_{2} = - 1 \to (- \frac{1}{2}) \times 2 = - 1 \end{array}$
General Form
- Parallel
  $\begin{array}{ll} 2 x - 3 y + 1 = 0 \\ 4 x - 6 y + 5 = 0 \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \to \frac{2}{4} = \frac{- 3}{- 6} \\ \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \to 0.5 \neq \frac{1}{5} & Not coincident \end{array}$
- Perpendicular
  $\begin{array}{ll} 3 x + 4 y - 1 = 0 \\ 4 x - 3 y + 2 = 0 \\ a_{1} a_{2} + b_{1} b_{2} = 0 \to 3 \times 4 + (4 \times - 3) = 0 \end{array}$

Jump to heading 10.Focus 4

Two lines are parallel

Analyze parallelism based on equal slopes, paying attention to the cases where the slope is $0$ or undefined.

Jump to heading $12$ Given that the line $l_{1} : (k - 3) x + (4 - k) y + 1 = 0$ is parallel to the line $l_{2} : 2 (k - 3) x - 2 y + 3 = 0$ , what is the value of $k ? .$

$\begin{array}{lllll} (A) 3 & (B) 5 & (C) 1 & (D) - 1 & (E) 3 \lor 5 \end{array}$

Jump to heading Solution

Substitute the options to verify the equation
$\begin{array}{ll} \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ \frac{k - 3}{2 (k - 3)} = \frac{4 - k}{- 2} \neq \frac{1}{3} \\ Pay attention to the special case where k = 3 \\ k = 3 {\begin{cases} \begin{array}{ll} l_{1} : y + 1 = 0 & Horizontal line a = 0 \\ l_{2} : - 2 y + 3 = 0 & Horizontal line a = 0 \end{array} \end{cases} \\ k = 3 ⟶ l_{1} and l_{2} are parallel \\ \frac{1}{2} = \frac{4 - k}{- 2} \neq \frac{1}{3} \\ Substitute k = 5 \\ \frac{1}{2} = \frac{4 - 5}{- 2} \\ \frac{1}{2} = \frac{1}{2} \neq \frac{1}{3} \end{array}$

Jump to heading Conclusion

Derived Solution
$(E)$
According to the Solution, get $k = 3, k = 5 l_{1} and l_{2} are parallel$ , so choose $E$ .
Formula used
$\begin{array}{ll} l_{1} ∥ l_{2} \Leftrightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} & Parallel positional relationship \\ a = 0 ⟶ b y + c = 0 & Horizontal line \end{array}$

Jump to heading 11.Focus 5

Two lines are perpendicular

When the product of the slopes of two lines is -1, or when their slopes are negative reciprocals of each other, the two lines are perpendicular. Note the special cases when the slope is 0 or undefined.

Jump to heading $13$ (Sufficiency judgment) Determine whether the condition that the lines $(m + 2) x + 3 m y + 1 = 0$ and $(m - 2) x + (m + 2) y - 3 = 0$ are perpendicular is sufficient.

$\begin{array}{lllll} (1) m = \frac{1}{2} . & (2) m = - 2. \\ (A) (1) & (B) (2) \\ (C) (1) + (2) & (D) (1), (2) \\ (E) \emptyset \end{array}$

$\begin{array}{ll} a_{1} a_{2} + b_{1} b_{2} = 0 \\ (m + 2) (m - 2) + 3 m (m + 2) = 0 \\ Verify condition (1) \\ (\frac{1}{2} + 2) (\frac{1}{2} - 2) + 3 \frac{1}{2} (\frac{1}{2} + 2) = 0 \\ (\frac{5}{2} \times - (\frac{3}{2})) + (3 \times \frac{5}{4}) = 0 \\ - \frac{15}{4} + \frac{15}{4} = 0 \\ Verify condition (2) \\ (- 2 + 2) (- 2 - 2) + 3 (- 2) (- 2 + 2) = 0 \\ (0 \times (- 4)) + (- 6 \times 0) = 0 \\ 0 + 0 = 0 \end{array}$

Jump to heading Conclusion

Derived Solution
$(D)$
According to the Solution, get $(1) ✅, (2) ✅$ , so choose $D$ .
Formula used
$\begin{array}{ll} a_{1} a_{2} + b_{1} b_{2} = 0 & Perpendicular positional relationship \end{array}$

Jump to heading $14$ If the line $m x + 3 y + 5 = 0$ is perpendicular to the line $n x - 2 y + 1 = 0$ , how many sets of positive integer solutions $(m, n)$ satisfy this condition $?$ .

$\begin{array}{lllll} (A) 1 & (B) 2 & (C) 3 & (D) 4 & (E) 5 \end{array}$

Jump to heading Solution

$\begin{array}{ll} a_{1} a_{2} + b_{1} b_{2} = 0 \\ m n + 3 \times (- 2) = 0 \\ m n = 6 \\ 6 = 1 \times 6 \\ 6 = 6 \times 1 \\ 6 = 2 \times 3 \\ 6 = 3 \times 2 \end{array}$

Jump to heading Conclusion

Derived Solution
$(D)$
According to the Solution, get $4 sets$ , so choose $D$ .

Formula used
$\begin{array}{ll} a_{1} a_{2} + b_{1} b_{2} = 0 & Perpendicular positional relationship \end{array}$

Jump to heading $15$ Given point $A (1, - 2)$ and point $B (m, 2)$ , and the equation of the perpendicular bisector of line segment $A B$ is $x + 2 y - 2 = 0$ , what is the value of the real number $m ? .$

$\begin{array}{lllll} (A) 3 & (B) 4 & (C) 5 & (D) 6 & (E) 7 \end{array}$

Jump to heading Solution

$1$ Solve by setting up equations
$\begin{array}{ll} (\frac{m + 1}{2}, 0) & Midpoint \\ \frac{m + 1}{2} + 0 - 2 = 0 \\ m + 1 + 0 - 2 = 2 \\ m + 1 = 4 \\ m = 3 \end{array}$

$2$ Solve using slopes
$\begin{array}{ll} k_{A B} = \frac{2 - (- 2)}{m - 1} \\ k = - \frac{a}{b} \\ x + 2 y - 2 = 0 ⟶ k = - \frac{1}{2} \\ The slope of A B and the slope of the line x + 2 y - 2 = 0 \\ are negative reciprocals of each other, since A B is perpendicular to x + 2 y - 2 = 0. \\ k_{A B} \Rightarrow \frac{2}{1} \\ k_{A B} = \frac{4}{m - 1} = 2 \\ 4 = 2 m - 2 \\ \frac{6}{2} = m \\ m = 3 \end{array}$

Jump to heading Conclusion

Derived Solution
$(A)$
According to the Solution, get $m = 3$ , so choose $A$ .
Formula used
$\begin{array}{ll} (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) & Midpoint formula \\ k = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} & Two-point slope formula \\ k = - \frac{a}{b} & Quickly convert the general form to the slope \end{array}$

Jump to heading $16$ Given that the equation of line $l$ is $x + 2 y - 4 = 0$ , and point $A$ has coordinates $(5, 7)$ , a line is drawn through point $A$ perpendicular to line $l$ . What is the x-coordinate of the foot of the perpendicular $? .$

$\begin{array}{lllll} (A) 6 & (B) 5 & (C) 2 & (D) - 2 & (E) - 1 \end{array}$

Jump to heading Solution

Show known conditions
$\begin{array}{ll} k = - \frac{a}{b} \\ l = x + 2 y - 4 = 0 ⟶ k = - \frac{1}{2} \end{array}$

$1$ Solve by using the two-point slope formula to write a system of equations and find the point of intersection
$\begin{array}{ll} Let B = (x_{0}, y_{0}) \\ k : l = - \frac{1}{2} ⟶ A B = \frac{2}{1} = 2 \\ {\begin{cases} \begin{array}{ll} x_{0} + 2 y_{0} - 4 = 0 & Point B on line l \\ \frac{y_{0} - 7}{x_{0} - 5} = 2 & The inclination from A to B \end{array} \end{cases} \\ x_{0} + 2 y_{0} - 4 = 0 & Simplify equation 1 \\ x_{0} = 4 - 2 y_{0} \\ \frac{y_{0} - 7}{x_{0} - 5} = 2 & Simplify equation 2 \\ y_{0} - 7 = 2 (x_{0} - 5) \\ y_{0} - 7 = 2 x_{0} - 10 \\ y_{0} = 2 x_{0} - 3 \\ y_{0} = 2 (4 - 2 y_{0}) - 3 & Substitute x_{0} \\ y_{0} = 8 - 4 y_{0} - 3 \\ y_{0} = 5 - 4 y_{0} \\ 5 y_{0} = 5 \\ y_{0} = 1 \\ x_{0} = 4 - 2 \times 1 & Substitute y_{0} \\ x_{0} = 2 \end{array}$

$2$ Solve by using the point-slope form to write a system of equations and find the point of intersection
$\begin{array}{ll} k : l = - \frac{1}{2} ⟶ A B = \frac{2}{1} = 2 \\ y = 2 (x - 5) + 7 = 2 x - 3 & The equation of the line A B \\ {\begin{cases} x + 2 y - 4 = 0 \\ y = 2 x - 3 \end{cases} \\ x + 2 (2 x - 3) - 4 = 0 & Substitute y \\ x + 4 x - 6 - 4 = 0 \\ 5 x - 10 = 0 \\ 5 x = 10 \\ x = 2 \end{array}$

Jump to heading Conclusion

Derived Solution
$(C)$
According to the Solution, get $x = 2$ , so choose $C$ .
Formula used
$\begin{array}{ll} k = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} & Two-point slope formula \\ k = - \frac{a}{b} & Quickly convert the general form to the slope \\ y = y_{0} + k (x - x_{0}) & Point-slope form \end{array}$

Jump to heading 12.Focus 6

Two lines intersect

When the slopes of two lines aren't equal, the lines intersect. Additionally, you should know how to find the coordinates of the intersection point.
Intersection point $=$ Solve the system of equations formed by the two lines.

Jump to heading $17$ (Sufficiency judgment) Determine whether the condition that the lines $(m + 1) x + 3 y + 1 = 0$ and $2 x + m y - 3 = 0$ intersect is sufficient.

$\begin{array}{lllll} (1) m > \frac{1}{2} . & (2) m < - 4. \\ (A) (1) & (B) (2) \\ (C) (1) + (2) & (D) (1), (2) \\ (E) \emptyset \end{array}$

Jump to heading Solution

$\begin{array}{ll} \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} & Intersect \to Not proportional \\ \frac{m + 1}{2} \neq \frac{3}{m} \\ m^{2} + m - 6 \neq 0 \\ m^{2} + m - 6 \Rightarrow (m - 2) (m + 3) = 0 \\ m = 2 \lor m = - 3 ⟶ m \neq 2 \land m \neq - 3 \\ m \in R ∖ {2, - 3} \\ Verify condition (1) \\ m > \frac{1}{2} \Rightarrow m = 2 \\ Verify condition (2) \\ m < - 4 \Rightarrow m \neq 2 \land m \neq - 3 \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $m < - 4 \Rightarrow m \neq 2 \land m \neq - 3$ , so choose $B$ .
Formula used
$\begin{array}{ll} \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} & Intersecting positional relationship \end{array}$

Jump to heading $18$ What is the distance from the intersection point of the lines $2 x + 3 y + 4 = 0$ and $3 x + y - 1 = 0$ to the origin $? .$

$\begin{array}{lllll} (A) \sqrt{2} & (B) \sqrt{3} & (C) 2 & (D) \sqrt{5} & (E) \sqrt{7} \end{array}$

Jump to heading Solution

$\begin{array}{ll} {\begin{cases} 2 x + 3 y + 4 = 0 \\ 3 x + y - 1 = 0 ⟶ 9 x + 3 y - 3 = 0 \end{cases} \\ (9 x + 3 y - 3) - (2 x + 3 y + 4) = 0 - 0 \\ 7 x - 7 = 0 \\ x = 1 \\ 3 \times 1 + y - 1 = 0 & Substitute x \\ y = - 2 \\ The distance from (1, - 2) to the origin. \\ d = \sqrt{(1 - 0)^{2} + (- 2 - 0)^{2}} = \sqrt{5} \end{array}$

Jump to heading Conclusion

Derived Solution
$(D)$
According to the Solution, get $d = \sqrt{5}$ , so choose $D$ .
Formula used
$\begin{array}{ll} d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} & Two-point distance formula \end{array}$

Jump to heading 13.Positional Relationship Between a Point and a Line

For a point $(x_{0}, y_{0})$ and a line $l : y = k x + b$
$y_{0} {\begin{cases} > k x_{0} + b, the point lies above the line. \\ = k x_{0} + b, the point lies on the line. \\ < k x_{0} + b, the point lies below the line. \end{cases}$

Jump to heading 14.Distance from a Point to a Line

For the line $l : a x + b y + c = 0,$ the distance from the point $(x_{0}, y_{0})$ to the line is $d = \frac{| a x_{0} + b y_{0} + c |}{\sqrt{a^{2} + b^{2}}} .$
Special case: The distance from $(0, 0)$ to the line $l$ is $d = \frac{| c |}{\sqrt{a^{2} + b^{2}}} .$

Jump to heading 15.Distance Between Two Parallel Lines

Given two parallel lines: $l_{1} : a x + b y + c_{1} = 0;$ $l_{2} : a x + b y + c_{2} = 0.$ The distance between $l_{1}$ and $l_{2}$ is $d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}} .$
Remark: The derivation process involves selecting an arbitrary point on one of the lines and then using the formula for the distance from a point to a line to calculate the distance between the two parallel lines.

Jump to heading 16.Focus 7

Positional Relationship Between a Point and a Line

First, convert the line into the form $y = k x + b$ , then substitute the point into the equation to make a determination.
Note: Make sure the coefficient of $y$ is positive; otherwise, the result of the judgment will be reversed.
General form: $a x + b y + c = 0$
$\begin{array}{ll} b > 0 {\begin{cases} \begin{array}{ll} a x + b y + c > 0 & The point is above the line. \\ a x + b y + c < 0 & The point is below the line. \end{array} \end{cases} \\ b < 0 {\begin{cases} \begin{array}{ll} a x + b y + c > 0 & The point is below the line. \\ a x + b y + c < 0 & The point is above the line. \end{array} \end{cases} \end{array}$

Jump to heading $19$ Given the equation of line $l : x + 2 y - 4 = 0,$ and the coordinates of point $A$ are $(5 - m, m) .$ If point $A$ is above the line $l,$ what is the range of values for $m ? .$

$\begin{array}{lllll} (A) m > 1 & (B) m > - 1 & (C) m > - 2 & (D) m > - \frac{1}{2} & (E) m < - 1 \end{array}$

Jump to heading Solution

$\begin{array}{ll} A : (5 - m, m) \\ 5 - m + 2 m - 4 > 0 \\ (5 - 4) + (- m + 2 m) > 0 \\ 1 + m > 0 \\ m > - 1 \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $m > - 1$ , so choose $B$ .
Formula used
$\begin{array}{ll} y = k x + b & Point-line position \end{array}$

Jump to heading 17.Focus 8

Distance from a Point to a Line

First, convert the line equation into general form, then apply the point-to-line distance formula.

Jump to heading $20$ Given point $C (2, - 3)$ , and points $M (5, 5), N (- 3, - 1)$ , what is the distance from point $C$ to the line $M N ? .$

$\begin{array}{lllll} (A) \frac{23}{5} & (B) \frac{22}{5} & (C) \frac{21}{5} & (D) \frac{19}{5} & (E) \frac{18}{5} \end{array}$

Jump to heading Solution

$\begin{array}{ll} K_{M N} = \frac{5 - (- 1)}{5 - (- 3)} = \frac{6}{8} = \frac{3}{4} \\ 3 x - 4 y + c = 0 & Reverse-derive the general formula for k = - \frac{a}{b} \\ 3 \times 5 - 4 \times 5 + c = 0 & Substitute M \lor N \\ 15 - 20 + c = 0 \\ c = 5 \\ 3 x - 4 y + 5 = 0 \\ 3 \times 2 - 4 \times (- 3) + 5 = 0 & Substitute C \\ d = \frac{| a x_{0} + b y_{0} + c |}{\sqrt{a^{2} + b^{2}}} \\ d = \frac{6 + 12 + 5}{\sqrt{3^{2} + 4^{2}}} = \frac{23}{5} \end{array}$

Jump to heading Conclusion

Derived Solution
$(A)$
According to the Solution, get $d = \frac{23}{5}$ , so choose $A$ .
Formula used
$\begin{array}{ll} k = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} & Two-point slope formula \\ k = - \frac{a}{b} & Quickly convert the general form to the slope \\ a x + b y + c = 0 & General form of a line \\ d = \frac{| a x_{0} + b y_{0} + c |}{\sqrt{a^{2} + b^{2}}} & Point-to-line distance formula \end{array}$

Jump to heading 18.Focus 9

Distance Between Two Parallel Lines

To apply the distance formula between two parallel lines, make sure to first unify the coefficients of $x$ and $y$ in both equations, and then proceed with the calculation.

Jump to heading $21$ Given $l_{1} : 3 x - 4 y + 2 = 0, l_{2} : 6 x - 8 y + 9 = 0$ , What is the distance between $l_{1}$ and $l_{2} ? .$

$\begin{array}{lllll} (A) \frac{7}{10} & (B) \frac{1}{4} & (C) \frac{1}{3} & (D) \frac{1}{2} & (E) \frac{7}{5} \end{array}$

Jump to heading Solution

Unify the $x$ and $y$ coefficients of the two parallel lines before solving
$\begin{array}{ll} l_{1} : 3 x - 4 y + 2 = 0 \to 6 x - 8 y + 4 = 0 & Multiply both sides by 2 \\ l_{2} : 6 x - 8 y + 9 = 0 \\ d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}} \\ d = \frac{| 9 - 4 |}{\sqrt{6^{2} + 8^{2}}} = \frac{5}{10} = \frac{1}{2} \end{array}$

Jump to heading Conclusion

Derived Solution
$(D)$
According to the Solution, get $d = \frac{1}{2}$ , so choose $D$ .
Formula used
$\begin{array}{ll} d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}} & Parallel Lines Distance Formula \end{array}$

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Jump to heading🔗 9.Positional Relationships Between Two Lines

Jump to heading🔗 10.Focus 4

Jump to heading🔗 12Given that the line l1:(k−3)x+(4−k)y+1=0 is parallel to the line l2:2(k−3)x−2y+3=0, what is the value of k?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 11.Focus 5

Jump to heading🔗 13(Sufficiency judgment) Determine whether the condition that the lines (m+2)x+3my+1=0 and (m−2)x+(m+2)y−3=0 are perpendicular is sufficient.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 14If the line mx+3y+5=0 is perpendicular to the line nx−2y+1=0, how many sets of positive integer solutions (m,n) satisfy this condition ?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 15Given point A(1,−2) and point B(m,2), and the equation of the perpendicular bisector of line segment AB is x+2y−2=0, what is the value of the real number m?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 16Given that the equation of line l is x+2y−4=0, and point A has coordinates (5,7), a line is drawn through point A perpendicular to line l. What is the x-coordinate of the foot of the perpendicular ?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 12.Focus 6

Jump to heading🔗 17(Sufficiency judgment) Determine whether the condition that the lines (m+1)x+3y+1=0 and 2x+my−3=0 intersect is sufficient.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 18What is the distance from the intersection point of the lines 2x+3y+4=0 and 3x+y−1=0 to the origin ?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 13.Positional Relationship Between a Point and a Line

Jump to heading🔗 14.Distance from a Point to a Line

Jump to heading🔗 15.Distance Between Two Parallel Lines

Jump to heading🔗 16.Focus 7

Jump to heading🔗 19Given the equation of line l:x+2y−4=0, and the coordinates of point A are (5−m,m). If point A is above the line l, what is the range of values for m?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 17.Focus 8

Jump to heading🔗 20Given point C(2,−3), and points M(5,5),N(−3,−1), what is the distance from point C to the line MN?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 18.Focus 9

Jump to heading🔗 21Given l1:3x−4y+2=0,l2:6x−8y+9=0, What is the distance between l1 and l2?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading 9.Positional Relationships Between Two Lines

Jump to heading 10.Focus 4

Jump to heading $12$ Given that the line $l_{1} : (k - 3) x + (4 - k) y + 1 = 0$ is parallel to the line $l_{2} : 2 (k - 3) x - 2 y + 3 = 0$ , what is the value of $k ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 11.Focus 5

Jump to heading $13$ (Sufficiency judgment) Determine whether the condition that the lines $(m + 2) x + 3 m y + 1 = 0$ and $(m - 2) x + (m + 2) y - 3 = 0$ are perpendicular is sufficient.

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $14$ If the line $m x + 3 y + 5 = 0$ is perpendicular to the line $n x - 2 y + 1 = 0$ , how many sets of positive integer solutions $(m, n)$ satisfy this condition $?$ .

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $15$ Given point $A (1, - 2)$ and point $B (m, 2)$ , and the equation of the perpendicular bisector of line segment $A B$ is $x + 2 y - 2 = 0$ , what is the value of the real number $m ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $16$ Given that the equation of line $l$ is $x + 2 y - 4 = 0$ , and point $A$ has coordinates $(5, 7)$ , a line is drawn through point $A$ perpendicular to line $l$ . What is the x-coordinate of the foot of the perpendicular $? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 12.Focus 6

Jump to heading $17$ (Sufficiency judgment) Determine whether the condition that the lines $(m + 1) x + 3 y + 1 = 0$ and $2 x + m y - 3 = 0$ intersect is sufficient.

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $18$ What is the distance from the intersection point of the lines $2 x + 3 y + 4 = 0$ and $3 x + y - 1 = 0$ to the origin $? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 13.Positional Relationship Between a Point and a Line

Jump to heading 14.Distance from a Point to a Line

Jump to heading 15.Distance Between Two Parallel Lines

Jump to heading 16.Focus 7

Jump to heading $19$ Given the equation of line $l : x + 2 y - 4 = 0,$ and the coordinates of point $A$ are $(5 - m, m) .$ If point $A$ is above the line $l,$ what is the range of values for $m ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 17.Focus 8

Jump to heading $20$ Given point $C (2, - 3)$ , and points $M (5, 5), N (- 3, - 1)$ , what is the distance from point $C$ to the line $M N ? .$

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 18.Focus 9

Jump to heading $21$ Given $l_{1} : 3 x - 4 y + 2 = 0, l_{2} : 6 x - 8 y + 9 = 0$ , What is the distance between $l_{1}$ and $l_{2} ? .$

Jump to heading Solution

Jump to heading Conclusion