Question

there are given the vertices of the triangle ABC :A(1;6).B(3;6).C(-4;3)......find the equation of the side BC

Answer 1

there are given the vertices of the triangle ABC :A(1;6).B(3;6).C(-4;3)......find the equation of the side BC

Answer 2

You need two components to form the equation of a line (the side BC). You need the gradient and the y-intercept.

Firstly work out the gradient, take point B and look at the change in x and compare it to the change in y, for every increase in x how much does the y value increase/decrease by? this is the gradient. Secondly once you have the gradient, take a point on the line and follow it via using the gradient to the value x=0 and read off the value of y and add it to your equation

this is now the equation of the line.

y=(gradient)x+(y intercept)

Answer 3

i.e. from B to C x decreases by 7 and decreases in y by 3 so the gradient is -3/-7=3/7

Answer 4

To calculate the y-intercept, do as follows:

Let C denote the y-intercept then

\[y= \frac{ 3 }{ 7 }x+C\]

Answer 5

Substitute in a value of x and y that you know is on the line i.e. B or C, lets do B, so subbing in (3,6):

6=\[\frac{ 3 }{ 7 } *3+C\] => 6-9/7=C => 33/7=C

Answer 6

B(3;6) . C (-4;3) =
x-x1/x2-x1=y-y1/y2-y1 =????

Answer 7

Hmm where have you got this from?