Question
Find the area of $\triangle\text{ABC}$ whose vertices are:
$A(1, 2), B(-2, 3)$ and $C(-3, -4)$
$A(1, 2), B(-2, 3)$ and $C(-3, -4)$
✓
Answer
$A(1, 2), B(-2, 3)$ and $C(-3, -4)$ are the vertices of
Then, $(x_1 = 1, y_1= 2), (x_2 = -2, y_2 = 3)$ and $(x_3 = -3, y_3 = -4)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{1(3-(-4))+(-2))(-4-2)+(-3)(2-3)\big\}$
$=\frac{1}{2}\big\{1(3+4)-2(-6)-3(-1)\big\}$
$=\frac{1}{2}\big\{7+12+3\big\}$
$=\frac{1}{2}(22)$
$=11\ \text{sq}.\text{units}$
Then, $(x_1 = 1, y_1= 2), (x_2 = -2, y_2 = 3)$ and $(x_3 = -3, y_3 = -4)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{1(3-(-4))+(-2))(-4-2)+(-3)(2-3)\big\}$
$=\frac{1}{2}\big\{1(3+4)-2(-6)-3(-1)\big\}$
$=\frac{1}{2}\big\{7+12+3\big\}$
$=\frac{1}{2}(22)$
$=11\ \text{sq}.\text{units}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.
→| Investment (Thousand Rupees) | $10-15$ | 15-20 | 20-25 | 25-30 | 30-35 |
| No. of families | 30 | 50 | 60 | 55 | 15 |
The sides of certain triangles are given below. Determine them are right triangles:
$(\text{a} - 1)\text{cm},2\sqrt{\text{a}}\text{ cm},(\text{a} + 1)\text{cm}.$
→$(\text{a} - 1)\text{cm},2\sqrt{\text{a}}\text{ cm},(\text{a} + 1)\text{cm}.$
Find the area of $\triangle\text{ABC}$ with $A(1, -4)$ and midpoints of sides through A being $(2, -1)$ and $(0, -1)$.
→If $12^{th}$ term of an A.P. is -$13$ and the sum of the first four terms is $24,$ what is the sum of first $10$ terms?
→In $\triangle ABC , \angle BAC =90^{\circ}$, seg $B L$ and seg $C M$ are medians of $\triangle A B C$. Then prove that: $4\left( BL ^2+C M^2\right)=5 B C^2$
→prove the theorem : In a right angled triangle, if the altitude is drawn to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other.
Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.
Show that the points (-3, -3), (3, 3) and $(-3\sqrt{3},3\sqrt{3})$ are the vertices of an equilateral triangle.
→Solve the following systems of equations:
$\sqrt{2}\text{x}-\sqrt{3}\text{y}=0,$
$\sqrt{3}\text{x}-\sqrt{8}\text{y}=0.$
→$\sqrt{2}\text{x}-\sqrt{3}\text{y}=0,$
$\sqrt{3}\text{x}-\sqrt{8}\text{y}=0.$
Find the area of the shaded region in the given figure, where a circular arc of radius 6cm has been drawn with vertex of an equilateral triangle of side 12cm as centre and a sector of circle of radius 6cm with centre B is made. $\big[\text{Use }\sqrt{3}=1.73\text{ and }\pi=3.14\big]$
→