If the vertices of a triangle are $(1, - 3), (4, p)$ and $(- 9, 7)$ and its' area is $15$ sq. units, find the value (s) of $p .$

Open in App

Solution

The vertices of the triangle are $A (1, - 3), B (4, p)$ and $C (- 9, 7) .$
Area of triangle $= 15$ sq.units.
Area of the triangle with vertices $(x_{1}, x_{2}); (x_{2}, y_{2}); (x_{3}, y_{3})$ is
$\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
Therefore,
$\frac{1}{2} [1. (p - 7) + 4 (7 + 3) - 9 (- 3 - p)] = | 15 |$
$p - 7 + 40 + 27 + 9 p = 2 \times | 15 |$
$10 p + 60 = \pm 30$
$p + 6 = \pm 3$
$p = - 3$ or $- 9.$

Suggest Corrections

Similar questions

Q. If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p. [CBSE 2012]

Q. (i) If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p.

(ii) Find the value of k so that the area of triangle ABC with A(k + 1, 1), B(4, –3) and C(7, –k) is 6 square units.

(i) If the vertices of $△ A B C$ be A(1, -3), B(4, p) and C(-9, 7) and its area is 15 square units, find the values of p.

(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is $(\frac{7}{2}, y)$ , find the value of y.

Join BYJU'S Learning Program

If the vertices of a triangle are (1,−3),(4,p) and (−9,7) and its' area is 15 sq. units, find the value (s) of p.

The vertices of the triangle are A(1,−3),B(4,p) and C(−9,7).Area of triangle =15 sq.units.Area of the triangle with vertices (x1,x2);(x2,y2);(x3,y3) is12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]Therefore, 12[1.(p−7)+4(7+3)−9(−3−p)]=|15|p−7+40+27+9p=2×|15|10p+60=±30p+6=±3p=−3 or −9.

If the vertices of a triangle are $(1, - 3), (4, p)$ and $(- 9, 7)$ and its' area is $15$ sq. units, find the value (s) of $p .$