Question

In figure, AB || PQ, AB = PQ, AC || PR and AC = PR, prove that BC || QR and BC = QR.​

Answer 1

Answer:

$\huge{\underline\mathfrak\red{Solution}}$

Given:

AB||PQ,AB = PQ,

• AC || PR and AC = PR,

To prove:

• BC||QR and BC=QR

Proof:

Since, in quadrilateral ABQP:-

AB||PQ and AB=PQ

=>Quadrilateral ABQP is parallelogram.

Therefore we can write:

AP||BQ and AP=BQ-------(i)

In quadrilateral APRC:-

AC||PR and AC=PR

=>Quadrilateral APRC is parallelogram.

Therefore we can write:

AP||CR and AP=CR-------(ii)

NOW,

From the equations i and ii, we get:

BQ||CR and BQ=CR

Hence proved!

Answer 2

Step-by-step explanation:

Let P and Q be the point of trisection of the line segment joining the points A (-5, 8) and B (10, −4).

So, AP = PQ = QB

We have AP : PB = 1 : 2 Co-ordinates of the point P are

\begin{gathered}\sf : \implies \bigg( \frac{1 \times 10 + 2 \times ( - 5) }{1 + 2} , \frac{1 \times ( - 4) + 2 \times 8}{1 + 2} \bigg) \\ \\ \sf : \implies \bigg( \frac{10 - 10 }{3} , \frac{16 - 4}{3} \bigg) \\ \\ \sf : \implies \bigg( \frac{0}{3} , \cancel{\frac{12}{3} } \bigg) \\ \\ \sf : \implies \bigg( 0, 4 \bigg)\end{gathered}

:⟹(

1+2

1×10+2×(−5)

,

1+2

1×(−4)+2×8

)

:⟹(

3

10−10

,

3

16−4

)

:⟹(

3

0

,

3

12

)

:⟹(0,4)

So point p lines on the y - axis

we have AQ : BQ = 2 : 1

Co-ordinates of the point Q are

\begin{gathered}\sf : \implies \bigg( \frac{2 \times 10 + 1 \times ( - 5) }{2 + 1} , \frac{2 \times ( - 4) + 2 \times 8}{2 + 1} \bigg) \\ \\ \sf : \implies \bigg( \frac{20 - 5 }{3} , \frac{ - 8 + 8}{3} \bigg) \\ \\ \sf : \implies \bigg( \cancel{ \frac{15}{3}} , \frac{ 0}{3} \bigg) \\ \\ \sf : \implies \bigg( 5, 0 \bigg) \\ \\\end{gathered}

:⟹(

2+1

2×10+1×(−5)

,

2+1

2×(−4)+2×8

)

:⟹(

3

20−5

,

3

−8+8

)

:⟹(

3

15

,

3

0

)

:⟹(5,0)