Question

12. Find a number t such that the line containing the points

Answer 1

Answer:

t= 11

Step-by-step explanation:

$\boxed{gradient = \frac{y1 - y2}{x1 - x2} }$

Gradient of line that contains points (3, 7) & (5, 11)

$= \frac{11 - 7}{5 - 3}$

$= \frac{4}{2}$

= 2

The product of the gradients of two perpendicular lines is -1.

Gradient of the line that contains points (t, -2) & (-3, 5)

= -1 ÷2

$= - \frac{1}{2}$

$\frac{ - 2 - 5}{t - ( -3 )} = - \frac{1}{2}$

$\frac{ - 7}{t + 3} = \frac{ - 1}{2}$

Cross multiply:

-(t +3)= -7(2)

Dividing by -1 on both sides:

t +3= 7(2)

t +3= 14

t= 14 -3

t= 11