Question

please give me the answers of question 5 ex 7.3 of class 10

Answer 1

Ques is the image

Answer- We have △ABC whose vertices are given.

We need to show that ar(△ABD) = ar(△ACD).

Using section formula to find coordinates of D, we get-
x=3+5/2=8/2=4
y=-2+2/2=0/2=0
Therefore, coordinates of point D are (4, 0)

Using formula to find area of triangle:

Area of △ABD =1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

= 1/2[4 (−2 − 0) + 3 {0 − (−6)} + 4 {−6 − (−2)}]

=1/2(−8 + 18 −16)

=1/2 (−6) = −3 sq units

Area cannot be in negative.

Therefore, we just consider its numerical value.

Therefore, area of △ABD = 3 sq units … (1)

Again using formula to find area of triangle:

Area of △ACD = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

=  1/2[4 (2 − 0) + 5 {0 − (−6)} + 4 {−6 −2 )}]

=1/2  (8 + 30 − 32) = ½ (6) = 3 sq units … (2)

From (1) and (2), we get ar(△ABD) = ar(△ACD)

Hence Proved.

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