BA ⊥ BD start overline, B, A, end overline, \perp, start overline, B, D, end overline \qquad m \angle CBD = 4x + 52^\circm∠CBD=4x+52 ∘ m, angle, C, B, D, equals, 4, x, plus, 52, degrees \qquad m \angle ABC = 8x - 10^\circm∠ABC=8x−10 ∘ m, angle, A, B, C, equals, 8, x, minus, 10, degrees Find m\angle CBDm∠CBDm, angle, C, B, D:

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)