Q) If Q(0,1) is equidistant from P(5,-3) and R (x,6), find the values of x. Ans: since the Point Q is equidistant, then Distance between P & Q = Distance between R & Q Â (5-0) 2 + (-3-1) 2 = (x-0)2 + (6-1)2 or 25 +16 = x2 + 25 or x2 = 16 […]
If Q(0,1) is equidistant from P(5,-3) and R (x,6), find the values of x. Read More »
Q) Show that the points (- 2, 3), (8, 3) and (6, 7) are the vertices of a right-angled triangle. Ans: Let’s plot these points on graph, we get: Step 1: Now for a Δ ABC to be an right angled triangle, required condition is: AB2 = AC2 + BC2 Step 2: Let’s calculate the
Q) Solve the pair of equations x = 5 and y = 7 graphically. Ans:Â Â Here is the answer to this question: For a detailed answer to this question, please refer to the similar question and its solution (click on the link below) Solve the pair of equations x = 3 and y =- 4
Solve the pair of equations x = 5 and y = 7 graphically. Read More »
Q) The centre of a circle is (2a, a-7). Find the values of ‘a’ if the circle passes through the point (11,-9). Radius of the circle is 5√2cm. Ans: Let’s find out radius of the circle by co-ordinate method. Therefore, (2a – 11)2 + (a -7 + 9)2 = (5√2 )2 or (4a2 + 121
Q) If (-5,3) coordinates and (5,3) are two vertices of an equilateral triangle, then find of the third vertex, given that origin lies inside the triangle. (Take √3 = 1.7) Ans:  VIDEO SOLUTION STEP BY STEP SOLUTION Let the third vertex be (x, y)Hence, 3 vertex of the triangle will be A (-5,3)         B (5,3)         C
Q) Solve the pair of equations x = 3 and y = – 4 graphically. Ans: VIDEO SOLUTION STEP BY STEP SOLUTION Step 1: Lets plot both the equations on the graph. Step 2: Increase till these lines intersect each other. We get the intersection point as (3,-4) Since the lines are intersecting at (3,
Solve the pair of equations x = 3 and y =- 4 graphically. Read More »