The triangle formed by the points (0, 7, 10), (-1, 6, 6), (-4, 9,… - Vidyadip

MCQ

The triangle formed by the points $(0, 7, 10), (-1, 6, 6), (-4, 9, 6)$ is

A
Equilateral
B
Isosceles
C
Right angled
✓
Right angled Isosceles

✓

Answer

Correct option: D.

Right angled Isosceles

d
(d) Let $A$ be $(0, 7, 10), B\ (-1, 6, 6)$ and $C\ (-4, 9, 6)$

Then, $AB = \sqrt {1 + 1 + 16} = \sqrt {18} = 3\sqrt 2 $

$BC = \sqrt {9 + 9 + 0} = \sqrt {18} = 3\sqrt 2 $

$AC = \sqrt {16 + 4 + 16} = \sqrt {36} = 6$

Clearly, $A{C^2} = A{B^2} + B{C^2}$

Hence $\Delta $ is right angled. Also, since $AB = BC$

$\Delta $ is right angled isosceles.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The minimum value of $f(a) = (2{a^2} - 3) + 3(3 - a) + 4$ is

For $x \in R$ , $f\left( x \right) = \left| {\log 2 - \sin x} \right|$ and $g\left( x \right) = f\left( {f\left( x \right)} \right)$ then .. .

Which of the following is incorrect

Let $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=\mathrm{fog}(10), \mathrm{b}=\operatorname{gof}(3)$. If $\mathrm{e}$ and $1$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$, then $8 e^2+1^2$ is equal to.

The value of a so that the sum of the squares of the roots of the equation ${x^2} - (a - 2)x - a + 1 = 0$ assume the least value, is

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b},$ then $\vec{a} \cdot \vec{b} \geq 0$ only when

The existence of the unique solution of the system $x + y + z = \lambda ,$ $5x - y + \mu z = 10$, $2x + 3y - z = 6$ depends on

If square root of, ${a^{\frac{1}{a}}}.{(2a)^{\frac{1}{{2a}}}}.{(4a)^{\frac{1}{{4a}}}}.{(8a)^{\frac{1}{{8a}}}}........\infty $ is $\frac{8}{{27}}$ , then the value of $'a'$ is

In a lottery there were $90$ tickets numbered $1$ to $90$. Five tickets were drawn at random. The probability that two of the tickets drawn numbers $15$ and $89$ is

The $20^{\text {th }}$ term from the end of the progression $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots .,-129 \frac{1}{4}$ is :-