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Answer 1

Answer:

first you have draw a number line on a graph paper

Step-by-step explanation:

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Answer 2

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Questions & Answers

CBSE

Mathematics

Grade 10

Distance Formula

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Find a point on the y-axis which is equidistant from the points A (6,5) and B (- 4,3)

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Hint: In this question remember that distance between the two points is given as; XY=(c−a)2+(d−b)2−−−−−−−−−−−−−−−√ here X is starting point whose coordinate is (a, b) and Y is the end point whose coordinate is (c, d), use this information to approach the solution.

Complete step-by-step solution:

Given points are A (6,5) and B (- 4,3)

Let us suppose the point on the y-axis which is equidistant from the given points be P (0, y) because as it lies on the y-axis so its x - coordinate will be 0

For point P to be equidistant from points A and B, distance between points A and P will be equal to the distance between points B and P

AP = BP (equation 1)

According to distance formula, the distance between any two points X (a, b) and Y (c, d) is given by

XY=(c−a)2+(d−b)2−−−−−−−−−−−−−−−√

Now substituting the values in the above equation, we get

AP=(0−6)2+(y−5)2−−−−−−−−−−−−−−−√=36+(y−5)2−−−−−−−−−−√

And BP=(0−(−4))2+(y−3)2−−−−−−−−−−−−−−−−−−√=16+(y−3)2−−−−−−−−−−√

Now substituting the values of AP and BP in equation 1 we get

36+(y−5)2−−−−−−−−−−√=16+(y−3)2−−−−−−−−−−√

Squaring the above equation both sides, we have

36+(y−5)2=16+(y−3)2

By applying the algebraic formula (a±b)2=a2+b2±2abin the above equation we get

36+y2+25−10y=16+y2+9−6y

⇒36+25−10y+6y−9−16=y2−y2

⇒36−4y=0

⇒4y=36

⇒y = 9

Therefore, the point on the y - axis which is equidistant from the points A (6, 5) and B (- 4, 3) is P (0, 9).

Note: In these types of problems, the statement of the problem should be carefully converted into an equation, which will be used to equate the given data and the unknown parameter which we need to find out. Here, if instead of the y-axis there would have been an x-axis so the coordinates of the point assumed would be (x, 0) instead of (0, y).