C1 Maths.. please help

The points A, B and C have coordinates (2,1), (-7,1), (5,4) respectively. The line through A perpendicular to the line BC intersects BC at the point D.

For part a) I have shown that the equation of BC is
x - 4y + 11 = 0
That's fine..

But how do I now find the equation of AD?
The gradient of AD must be -4 but what other values should I plug into y-y1=m(x-x1)

Please do explain.
Thanks in advance :)

m of AD = -4
you have a point on the line AD which is A(2,1)
using the point A, m = -4, and the equation y=mx+c
1 = -4(2) + c
9 = c

equation of AD: y = -4x + 9

You know that AD is Perpendicular to BC

And when its perpendicular the gradient is switched upside down and the sign becomes the opposite.

So it would be : Y = -4x - 9



By the way, You should put the equation is y = mx +c form.

first of all, happy new year wishes...

You have found
Eqn of BC=x-4y+11=0.

In general,if ax+by+c=0,
slope=(-a/b)=(-1/-4)=1/4

Slope of perpendicular line AD=-1/m
=-1/(1/4)
=-4
Eqn of AD
y=mx+c
y=-4x+c

It passes through A(2,1)
1=-4(2)+c
c=1+8=9

So, eqn of AD
y=-4x+9


Hope it helps