Find the equation of the line that passes through the point $(4, 5)$ and is perpendicular to the line $3 x - 4 y + 2 = 0$ . Then find its point of intersection with the line $2 x + y - 7 = 0$ .

medium 3 marks Class 10 · Mathematics · Coordinate Geometry

Answer

(-5,17)

The correct answer is (-5,17) — This completes the problem.

Step-by-step solution

Write the given line in slope–intercept form: $3 x - 4 y + 2 = 0 ⟹ y = \frac{3}{4} x + \frac{1}{2}$ , so its slope is $m_{1} = \frac{3}{4}$ .
Perpendicular slopes are negative reciprocals. Hence the required line has slope $m_{2} = - \frac{4}{3}$ .
Use point–slope form with point $(4, 5)$ : $y - 5 = - \frac{4}{3} (x - 4)$ .
Simplify: $y = - \frac{4}{3} x + \frac{31}{3}$ .
Convert to standard form: $4 x + 3 y - 31 = 0$ .
For the second line, rewrite $2 x + y - 7 = 0$ as $y = 7 - 2 x$ .
Substitute $y$ into the first line: $4 x + 3 (7 - 2 x) - 31 = 0$ .
Simplify: $4 x + 21 - 6 x - 31 = 0 im pl i es - 2 x - 10 = 0 im pl i es x = - 5$ .
Find $y$ using $y = 7 - 2 x$ : $y = 7 - 2 (- 5) = 17$ .
So the intersection point is $(- 5, 17)$ .
This completes the problem.

Common mistake: Forgetting the negative reciprocal of the slope

Concept tested: Equation of perpendicular line