Question
Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.
✓
Answer
In a right angles triangle ABC, right angled at B, according to the pythagoras theorem
$\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2$
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}\dots(1)$
For the given points Distance between P and Q is
$P Q=\sqrt{(-2-2)^2+(2-2)^2}=\sqrt{16} $
$Q R=\sqrt{(2-2)^2+(7-2)^2}=\sqrt{25} $
$P R=\sqrt{(-2-2)^2+(2-7)^2}=\sqrt{16+25}=\sqrt{41}$
$\mathrm{PQ}^2=16$
$\mathrm{QR}^2=25$
$\mathrm{PR}^2=41$
As $P Q^2+Q^2=P R^2$
Hence the given points form a right angled triangle.
$\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2$
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}\dots(1)$
For the given points Distance between P and Q is
$P Q=\sqrt{(-2-2)^2+(2-2)^2}=\sqrt{16} $
$Q R=\sqrt{(2-2)^2+(7-2)^2}=\sqrt{25} $
$P R=\sqrt{(-2-2)^2+(2-7)^2}=\sqrt{16+25}=\sqrt{41}$
$\mathrm{PQ}^2=16$
$\mathrm{QR}^2=25$
$\mathrm{PR}^2=41$
As $P Q^2+Q^2=P R^2$
Hence the given points form a right angled triangle.
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 given figure, there are two concentric circles with centre O of radii $5\ cm$ and $3\ cm$. From an external point P, tangent PA and PB are drawn to these circles.
If $AP = 12\ cm$, find the length of BP.
→If $AP = 12\ cm$, find the length of BP.
Write the value of $\lambda,$ for which $\text{x}^2+4\text{x}+\lambda$ is a perfect square.
→An archery target has three regions formed by the concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions.
Show that the system of equations:
$\text{6x}+\text{5y}=11,\ \text{9x}+\frac{15}{2}\text{y}=21$
has no solutions.
→$\text{6x}+\text{5y}=11,\ \text{9x}+\frac{15}{2}\text{y}=21$
has no solutions.
If $\triangle\text{ABC}$ is a right triangle such that $\angle\text{C} = 90^\circ,\angle\text{A}=45^\circ$ and BC = 7 units. Find $\angle\text{B}$ AB and AC.
→Find the ratio in which the point P(m, 6) divides the join of A(-4, 3) and B(2, 8). Also find the value of m.
The wheels of a car make 2500 revolutions in covering a distance of 4.95km. Find the diameter of a wheel.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→Prove : $\left(1+\cot ^2 \theta\right)(1+\cos \theta)(1-\cos \theta)=1$
→Solve the following simultaneous equation.
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
→$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio.
