Finding the coordinate C.

Question

A triangle $A$, $B$, $C$ has the coordinates:

$A = (-1, 3)$ $B = (3, 1)$ $C = (x, y)$

$BC$ is perpendicular to $AB$. Find the coordinates of $C$

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My attempt:

Grad of $AB$ =

$$\frac{3-1}{-1-3} = -0.5$$

Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular).

Equation of $BC$

$(y-1) = 2(x-3)$

$y = 2x - 5$

Equation of $AC$

$(y-3) = m(x--1)$

$y = mx+m+3$

I do not know how to proceed further. Please help me out.

Answer 1

The equation of $AC$ is $x-3y=-10$ as slope of any line is $-\frac{a}{b}$ where $a$ is $x$-coordinate and $b$ is $y$-coordinate so slope is $-\left(\frac{-1}{3}\right)$. $C$ is the point where $AC$ and $BC$ so meet we have two simultaneous equations $2x-y=5$ and $x-3y=-10$ solving them you get $x=5$ and $y=5$.

Answer 2

HINT:

Length of $AC =$

$\sqrt{((x+1)^2 +(y-3)^2)} = \sqrt{(\mathrm{Length} \space \mathrm{of} \space (AB)^2+\mathrm{Length} \space \mathrm{of} \space (BC)^2)}\tag{1}$

because you have a right-angled triangle as $BC$ is perpendicular to $AB$

From $(1)$ $$(x+1)^2 +(y-3)^2=(3+1)^2+(1-3)^2+(x-3)^2+(y-1)^2$$ $$\implies x^2+2x+1 +y^2-6y+9=20+x^2-6x+9+y^2-2y+1$$

Answer 3

Yes, the gradient of line AB is $-1/2$ so the gradient of any line perpendicular to $AB$ is $2$. The line with gradient $2$ through $B = (3, 1)$ is $y = 2(x- 3)+ 1$ or $y= 2x- 5$. All points on that line, $(x, 2x- 5)$, satisfy the conditions for C.