As per the question given: \[tan\text{ }A\text{ }=\text{ }5/6\text{ }and\text{ }tan\text{ }B\text{ }=\text{ }1/11\] Since,\[tan\text{ }\left( A\text{ }+\text{ }B \right)\text{ }=\text{ }\left(...
Solution: (i) Let (p, q) be the coordinates of a point Q. Provided, The point Q (p, q), Divide the line joining $\mathrm{B}\left(\mathrm{x}{2}, \mathrm{y}{2}\right)$ and...
Solution: According to the given question, A, B and C are the vertices of ΞABC A(x1, y1), B(x2, y2), C(x3, y3) are the coordinates of A, B and C. (i) According to the information provided, D is BC's...
Solution: According to the given question, A (6, 1), B (8, 2) and C (9, 4) are the three vertices of a parallelogram ABCD Let (x, y) be the fourth vertex of parallelogram. It is known to us that,...
Solution: Let (x,y) be the vertices. The distance between (x,y) & (4,3) is $=\text{ }\surd ({{\left( x-4 \right)}^{2}}~+\text{ }{{\left( y-3 \right)}^{2}})$β¦β¦(1) The distance between (x,y) &...
Solution: Let 1: k be the ratio in which x-axis divides the line segment joining (β4, β6) and (β1, 7). Therefore, x-coordinate is (-1 β 4k) / (k + 1) y-coordinate is (7 β 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(β2, β3) and C(8, 2m) are collinear. Therefore the area of βABC = 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 (β5, 4) and B (β1, 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 (β3, β14) and B (a, β5): $=\surd...
Solution: A (2, β2), B (7, 3), C (11, β1) and D (6, β6) 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 (β5, 6), B (β4, β2) 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{...
Solution: The statement given in the question is false. Justification: A (4, 3), B (6, 4), C (5, β6) and D (β3, 5) are the points given. We need to find the distance between A and B...
Solution: The statement given in the question is true. Justification: Coordinates of A $=\text{ }({{x}_{1}},\text{ }{{y}_{1}})\text{ }=\text{ }\left( 3,\text{ }1 \right)$ Coordinates of B $=\text{...
Solution: The statement given in the question is false. Justification: We know that the points on the perpendicular bisector of the line segment joining two points are equidistant from the two...
Solution: Option (C) 10 is the correct answer. The formula for distance: ${{d}^{2}}~=\text{ }{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}$ According to the...
Solution: Option (C) 8 is the correct answer. The vertices of the triangle are, $A\text{ }({{x}_{1}},\text{ }{{y}_{1}})=\left( 3,\text{ }0 \right)$ $B\text{ }({{x}_{2}},\text{ }{{y}_{2}})=\left(...
Solution: Option (B) 12 is the correct answer. (0, 4), (0, 0) and (3, 0) are the vertices of a triangle. The perimeter of triangle AOB = Sum of the length of all its sides: = distance between...
Solution: Option (C)Β β34 is the correct answer. The three vertices are: $A\text{ }=\text{ }\left( 0,\text{ }3 \right)$, $O\text{ }=\text{ }\left( 0,\text{ }0 \right)$ , $B\text{ }=\text{ }\left(...
Solution: Option (B)Β 5β 2 is the correct answer. Distance formula: ${{d}^{2}}~=\text{ }{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}$ According to the given...
Solution: Option (B) 8 is the correct answer. The formula for distance : ${{d}^{2}}~=\text{ }{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}$ According to the...
Solution: Option (B) 3 is the correct answer. We all know that, On the Cartesian plane (x, y) is a point in first quadrant. Then, Perpendicular distance from Yβaxis = x, and Perpendicular distance...