math question please help?

Question

if the graphs of the equations 3x-5y+4=0 and 2x+ay-11=0 intersect at right angles, find the value of a.

Answer 1

If two equations intersect at right angles, the multiple of the slopes of the equations must equate to -1. Therefore, the slope of:

1st equation: 3/5
2nd equation: -2/a
=> (3/5)*(-2/a)=-1, and a = 1.2


XR

Answer 2

If they intersect at right angles, it means they are perpendicular. Perpendicular lines have opposite, flipped slopes. So,

Find the slope of 3x-5y+4=0 ---put in slope-intercept form (y=mx+b)


3x-5y=-4 We are solving for y

-5y= -3x-4 Divide ALL terms by -5

y=3/5x+4/5
y=mx+b the slope for this line is 3/5

that means the slope for 2x+ay-11=0 is -5/3

2x+ay-11=0 solve for y

2x+ay=11

ay= -2x +11

divide ALL terms by a

y= -2x/a +11/a

-2/a=-5/3 cross multiply

-6= -5a

a= 6/5

Answer 3

If they intersect at right angles, their slopes are perpendicular and the product of their slopes would be -1.

3x - 5y = 4
-5y = -3x + 4
y = 3/5 x - 4/5

slope = 3/5

2x + ay = 11
ay = -2x + 11
y = -2/a x + 11/a

slope = -2/a

But we know that (3/5)(-2/a)= -1, so,
-6/(5a) = -1
a = 6/5

that's it! :)