With reference to the conventional cartesian $(x, y)$ coordinate system, the vertices of a triangle have the following coordinates;
$(x_{1}, y_{1}) = (3, 2); (x_{2}, y_{2}) = (4, 4) a n d (x_{3}, y_{3}) = (6, 5)$ .
The area of the triangle is equal to

$\frac{3}{4}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{5}{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{2}{3}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{3}{2}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$\frac{3}{2}$

Area of triangle $= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | 3 (4 - 5) + 4 (5 - 2) + 6 (2 - 4) |$

$= \frac{1}{2} | - 3 + 12 - 12 | = \frac{3}{2}$

Suggest Corrections

Similar questions

Q. With referance to the conventional certesian

(x, y)

coordinate system, the vertices of the triangle have the following coordinates :

(x_{1}, y_{1})

(1, 0)

;

(x_{2}, y_{2})

(2, 2)

and

(x_{3}, y_{3})

(4, 3)

. The area of the triangle is equal to

ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

Q. If

A (x_{1}, y_{1}), B x_{2}, y_{2}), C (x_{3}, y_{3})

are the vertices of an equilateral triangle and

(x_{1} - 2)^{2} + (y_{1} - 3)^{2} = (x_{2} - 2)^{2} + (y_{2} - 3)^{2} = (x_{3} - 2)^{2} + (y_{3} - 3)^{2}

, then

Prove that the coordinates of the centroid of the triangle whose vertices are $(x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3})$ are $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ and also, deduce that the medians of a triangles are concurrent.

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

Join BYJU'S Learning Program

With reference to the conventional cartesian(x,y) coordinate system, the vertices of a triangle have the following coordinates; (x1,y1)=(3,2);(x2,y2)=(4,4) and (x3,y3)=(6,5). The area of the triangle is equal to

The correct option is D 32 Area of triangle=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)| =12|3(4−5)+4(5−2)+6(2−4)| =12|−3+12−12|=32

With reference to the conventional cartesian $(x, y)$ coordinate system, the vertices of a triangle have the following coordinates;
$(x_{1}, y_{1}) = (3, 2); (x_{2}, y_{2}) = (4, 4) a n d (x_{3}, y_{3}) = (6, 5)$ .
The area of the triangle is equal to

The correct option is D
$\frac{3}{2}$

Area of triangle $= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | 3 (4 - 5) + 4 (5 - 2) + 6 (2 - 4) |$

$= \frac{1}{2} | - 3 + 12 - 12 | = \frac{3}{2}$