(a) (4,4) is a point on a locus whose equation is y2 = kx. Show that (16,8) is another point on the locus. (b) (2,3) is a point on a locus whose equation is 5x +by= 16. Show that (0,8) is another point of the locus

Please check the explanation.

Step-by-step explanation:

a)

Given

(4,4) is a point on a locus whose equation is y²=kx

Putting (4, 4) in y²=kx

(4)² = k(4)

16 = 4k

k = 16/4

k = 4

Checking the point (16, 8) on the locus

y²=kx

Putting (16, 8) and k =4 in y²=kx

y²=kx

(4)² = 4×4

16 = 16

L.H.S = R.H.S

Therefore, (16, 8) is another point on the locus.

b)

Given

(2, 3) is a point on a locus whose equation is 5x +by= 16

Putting (2, 3) in 5x +by= 16

5x+by=16

5(2)+b(3)=16

10+3b=16

3b=6

b=6/3

b=2

Putting (0, 8) and b=2 in 5x +by= 16

5x +by= 16

5(0)+2(8)=16

0+16=16

16=16

L.H.S = R.H.S

Therefore, (0, 8) is another point on the locus.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.