A line $L$ passes through the point $A(3,-2)$ and is perpendicular to the line $4x+3y-7=0$.

Step-by-step solution

1. Rewrite $4x+3y-7=0$ as $y=-\frac{4}{3}x+\frac{7}{3}$, giving slope $m_1=-\frac{4}{3}$.
2. For a line perpendicular to this, the slope is the negative reciprocal: $m_2=\frac{3}{4}$.
3. Use point‑slope form with point $A(3,-2)$: $y-(-2)=\frac{3}{4}(x-3)$.
4. Simplify: $y+2=\frac{3}{4}x-\frac{9}{4}$.
5. Multiply by 4: $4y+8=3x-9$.
6. Bring all terms to one side: $3x-4y-17=0$. This is the equation of $L$.
7. Now to find the foot of the perpendicular from $B(7,4)$ to $L$, write $L$ as $ax+by+c=0$ with $a=3$, $b=-4$, $c=-17$.
8. Compute $D=a{x}_0+b{y}_0+c=3\cdot 7+(-4)\cdot 4-17=21-16-17=-12$.
9. Calculate $a^2+b^2=3^2+(-4{)}^2=9+16=25$.
10. Find $x^{\prime }=x_0-a\frac{D}{a^2+b^2}=7-3\left(\frac{-12}{25}\right)=7+\frac{36}{25}=\frac{211}{25}$.
11. Find $y^{\prime }=y_0-b\frac{D}{a^2+b^2}=4-(-4)\left(\frac{-12}{25}\right)=4-\frac{48}{25}=\frac{52}{25}$.
12. Thus the foot of the perpendicular is $(\frac{211}{25},\frac{52}{25})$.

Concept tested: Perpendicular lines and foot of a perpendicular (using formula)

Published 20 August 2026 · How this solution was prepared and checked