Solution: Let 1: k be the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7). Therefore, x-coordinate is (-1 – 4k) / (k + 1) y-coordinate is (7 – 6k) / (k + 1) y coordinate...

Solution: The provided vertices are: $({{x}_{1}},\text{ }{{y}_{1}})\text{ }=\text{ }\left( -8,\text{ }4 \right)$ $({{x}_{2}},\text{ }{{y}_{2}})\text{ }=\text{ }\left( -6,\text{ }6 \right)$...

Solution: $A(2,-4), P(3,8)$ and $Q(-10, y)$ are the given points. Now according to the question given, $$ \begin{aligned} P A &=Q A \\ \sqrt{(2-3)^{2}+(-4-8)^{2}} &=\sqrt{(2+10)^{2}+(-4-y)^{2}} \\...

Solution: The points given here i.e., A(5, 1), B(–2, –3) and C(8, 2m) are collinear. Therefore the area of ∆ABC = 0 ${\scriptscriptstyle 1\!/\!{ }_2}\text{ }[{{x}_{1}}~({{y}_{2}}~\text{...

Solution: As the point P lies on the perpendicular bisector of AB, point Q is the midpoint of AB . By the formula for midpoint: $({{x}_{1}}~+\text{ }{{x}_{2}})/2\text{ }=\text{ }\left( -5+4...

Solution: Let P be the point. Now according to the given question, P is at equal distance from A (–5, 4) and B (–1, 6) Then the point P $=\text{ }(({{x}_{1}}+{{x}_{2}})/2,\text{...

Solution: The distance between the two points (x1,y1) ( x2,y2) : $d=\surd ({{x}_{2}}-{{x}_{1}}){}^\text{2}+({{y}_{2}}-{{y}_{1}}){}^\text{2}$ The distance between A (–3, –14) and B (a, –5): $=\surd...

Solution: A (2, –2), B (7, 3), C (11, –1) and D (6, –6) are the given points. Now using the distance formula, $d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{...

Solution: (x, 0) = Let coordinates of the point (given that the point lies on x axis) ${{x}_{1}}=7.\text{ }{{y}_{1}}=-4$ ${{x}_{2}}=x.\text{ }{{y}_{2}}=0$ Distance $=\surd...

Solution: A (–5, 6), B (–4, –2) and C (7, 5) are the given points. Now, using the distance formula, $d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{...