Question

find the values of k such that area of triangle pqr is 6 square units when coordinates of p( k + 1,1) ,Q(4,-3) and R(7,-3)

Answer 1

In the above Question , the following information is given -

A triangle , ∆ PQR whose coordinates are P ( k + 1, 1 ) , Q ( 4, -3 ) and R ( 7, -3 ) has the area of 6 square units .

To find -

We have to find the value of k .

Solution -

∆ PQR -

Coordinates of ∆ PQR -

P - [ k + 1, 1 ]

Q - [ 4, -3 ]

R - [ 7, -3 ]

P - [ k + 1, 1 ]

Area -

½ | ( k + 1 ) × ( -3 ) + ( 4 × -3 ) + ( 7 × 1 ) - ( 1 × 4 ) + ( 3 × 7 ) + { 3k + 3 } |

=> ½ | - 3k - 3 - 12 + 7 - 4 + 21 + 3k + 3 |

=> ½ | 12 |

=> 6 .

Thus , we can say that the k component get cancelled.

So , the value of k is 0.

This is the required answer.

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