Question

Find the equation of a straight line which passes through the point (3, 2) and is perpendicular to the line 3x + 5y =10. Hence, determine the gradient and y-intercept of the equation.​

Answer 1

Explanation:

eq of line perpendicular to 3x+5y=10 is

5x-3y=a

where a is constant.

it passes through (3,2)

5(3)-3(2)=a

a=15-6=9

reqd. eq. is 5x-3y=9

or

3y=5x-9

y=5/3 x-3

gradient=5/3

y-intercept=-3

or

5y=-3x-10

y=-3/5 x-2

slope=-3/5

slope of reqd. line=-1/(-3/5)=5/3

eq. of line through (3,2) is

y-2=5/3(x-3)

3y-6=5x-15

3y=5x-15+6

3y=5x-9

y=5/3 x-3