**Question 1:**

**Points A(-1, y), B(5, 7) lie on a circle with centre O(2, -3y). Find the values of y.**

**Solution:**

The given points are A(-1, y), B(5, 7) and O(2, -3y).

Here, AO and BO are the radii of the circle. So

AO = BO

Hence, AO^{2} = BO^{2}

\( \Rightarrow \) = (2 + 1)^{2} + (-3y – y)^{2} = (2 – 5)^{2} + (-3y – 7)^{2}

\( \Rightarrow \) 9 + (4y)^{2} = (-3)^{2} + (3y + 7)^{2 }

\( \Rightarrow \) 9 + 16y^{2} = 9 + 9y^{2} + 49 + 42y

\( \Rightarrow \) 7y^{2} – 42y – 49 = 0

\( \Rightarrow \) y^{2} – 6y – 7 = 0

\( \Rightarrow \) y^{2} – 7y + y – 7 = 0

\( \Rightarrow \) y(y – 7) + 1(y – 7) = 0

\( \Rightarrow \) (y – 7) (y + 1) = 0

\( \Rightarrow \) y = -1 or y = 7

Hence, y = 7 or y = -1

**Question 2:**

**If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find P.**

**Solution:**

The given points are A(0, 2), B(3, p) and C(p, 5)

AB = AC \( \Rightarrow \) AB^{2} = AC^{2}

\( \Rightarrow \) (3 – 0)^{2} + (p – 2)^{2} = (p – 0)^{2} + (5 – 2)^{2}

\( \Rightarrow \) 9 + p^{2} – 4p + 4 = p^{2} + 9

\( \Rightarrow \) 4p = 4 \( \Rightarrow \) p = 1

Hence, p = 1.

**Question 3:**

**ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Find the length of one of its diagonal.**

The given vertices are B(4, 0) , C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So,

\( BD = \sqrt{(4 – 0)^{2} + (0 – 3)^{2}} \)

\( \Rightarrow \sqrt{4^{2} + (-3)^{2}} \)

\( \Rightarrow \sqrt{16 + 9} \)

\( \Rightarrow = \sqrt{25} = 5 \)

Hence, the length of the diagonal is 5 units

**Question 4:**

**If the point P(k-1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.**

The given points are P(k-1, 2), A(3, k) and B(k, 5)

AP = BP \( \Rightarrow \) AP^{2} = BP^{2}

\( \Rightarrow \) (k -1 – 3)^{2} + (2 – k)^{2} = (k – 1 – k)^{2} + (2 – 5)^{2}

\( \Rightarrow \) (k – 4)^{2} + (2 – k)^{2 }= (-1)^{2} + (-3)^{2}

\( \Rightarrow \) k^{2} – 8y + 16 + 4 + k^{2} – 4k = 1+ 9

\( \Rightarrow \) k^{2} – 6y + 5 = 0

\( \Rightarrow \) (k – 1) (k – 5) = 0

\( \Rightarrow \) k = 1 or k = 5

Hence, k = 1 or k = 5.

**Question 5:**

**Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, -3).**

**Solution:**

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x_{1} = 12, y_{1} = 5) and B(x_{2} = 4, y_{2} = -3) then,

\( x = \frac{k \times 4 + 12}{k + 1} \) and \( 2 = \frac{k \times (-3) + 5}{k + 1} \)

Now,

\( 2 = \frac{k \times (-3) + 5 }{k + 1} \Rightarrow 2k + 2 = -3k + 5 \Rightarrow k = \frac{3}{5} \)

Hence, the required ration is 3:5.

**Question 6:**

**Prove that the diagonals of a rectangle ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6) are equal and bisect each other.**

The vertices of the rectangle ABCD are A(2, -1) , B(5, -1), C(5, 6) and D(2, 6). Now,

Coordinates of midpoint of AC = \( \frac{2 + 5}{2}, \frac{-1 + 6}{2} = \frac{7}{2}, \frac{5}{2} \)

Coordinates of midpoint of BD = \( \frac{5 + 2}{2}, \frac{-1 + 6}{2} = \frac{7}{2}, \frac{5}{2} \)

Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

**Question 7:**

**Find the lengths of the medians AD and BE of triangle ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).**

The given vertices are A(7, -3), B(5, 3) and C(3, -1)

Since D and E are the midpoints of BC and AC respectively, therefore

Coordinates of D = \(\frac{5 + 3}{2} , \frac{3 – 1}{2} = (4, 1)\)

Coordinates of E = \(\frac{7 + 3}{2} , \frac{-3 -1}{2} = (5, -2)\)

Now,

\( AD = \sqrt{(7 – 4)^{2} + (-3 -1)^{2}} = \sqrt{9 + 16} = 5 \)

\( BE = \sqrt{(5 – 5)^{2} + (3 + 2)^{2}} = \sqrt{0 + 25} = 5 \)

Hence, AD = BE = 5 units

**Question 8:**

**If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3 then find the value of K.**

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3. So,

\( k = \frac{2 \times 5 + 3 \times 2}{2 + 3} \)

= \( \frac{10 + 6}{5} \)

\( = \frac{16}{5} \)

Hence, \( k = \frac{16}{5} \).

**Question 9:**

**Find the point on x-axis which is equidistant from points A(-1, 0) and B(5, 0).**

Let p(x, 0) be the point on x-axis. Then,

AP = BP \( \Rightarrow \) AP^{2} = BP^{2}

\( \Rightarrow \) (x + 1)^{2 Â }+ (0 – 0)^{2} = (x – 5)^{2} + (0 – 0)^{2}

\( \Rightarrow \) x^{2} + 2x + 1 = x^{2} – 10x + 25

\( \Rightarrow \) 12x = 24 \( \Rightarrow \) x = 2

Hence, x = 2.

**Question 10:**

**Find the distance between the points \( A(\frac{-8}{5}, 2) and B(\frac{2}{5}, 2) \)**

The given points are \( A(\frac{-8}{5}, 2) and B(\frac{2}{5}, 2) \)

Then, \( (x_{1} = \frac{-8}{5}, y_{1} = 2) and (x_{2} = \frac{2}{5}, y_{2} = 2) \)

Therefore,

\( AB = \sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}} \)

= \( \sqrt{{\frac{2}{5} – \frac{-8}{5}}^{2} + (2 – 2)^{2}} \)

= \( \sqrt{2^{2} + 0^{2}} \)

= \( \sqrt{4 + 0} \)

= \( \sqrt{4} = 2 \, units \)

**Question 11:**

**Find the value of a, so that the point (3, a) lies on the line represented by 2x-3y = 5.**

The point (3, a) lies on the line 2x -3y = 5

If point (3, a) lies on the line 2x -3y = 5, then

2x â€“ 3y = 5

\(\Rightarrow\) (2*3) â€“ (3*a) = 5

\(\Rightarrow\) 6 â€“ 3a = 5

\(\Rightarrow a=\frac{1}{3}\)

Hence, the value of a is \(a=\frac{1}{3}\)

**Question 12:**

**If the points A(4,3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.**

The given points A (4, 3) and B (x, 5) lie on the circle with center O (2, 3)

Then, OA = OB

\(\Rightarrow \sqrt{(x-2)^{2}+(5-3)^{2}} = \sqrt{(4-2)^{2}+(3-3)^{2}}\)

\(\Rightarrow (x-2)^{2} + 2^{2} = 2^{2} + 0^{2}\)

\(\Rightarrow (x-2)^{2} = (2^{2} – 2^{2})\)

\(\Rightarrow (x-2)^{2} = 0\)

\(\Rightarrow (x-2) = 0\)

\(\Rightarrow x = 2\)

Hence, the value of x = 2

**Question 13:**

**If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.**

Let the point P (x, y) be equidistant from the points A (7, 1) and B (3, 5)

Then, PA = PB

\(\Rightarrow PA^{2}=PB^{2}\)

\(\Rightarrow \sqrt{(x-7)^{2}+(y-1)^{2}} = \sqrt{(x-3)^{2}+(y-5)^{2}}\)

\(\Rightarrow x^{2} + y^{2} – 14x – 2y + 50 = x^{2} +y^{2} -6x -10y +34\)

\(\Rightarrow 8x â€“ 8y = 16\)

\(\Rightarrow x â€“ y = 2\)

**Question 14:**

**If the centroid of triangle ABC having vertices A(a, b), B(b, c) and C(c, a) is the origin, then find the value of (a + b + c).**

The given points are A (a, b), B (b, c) and C (c, a)

Here,

(x_{1} = a, y_{1} = b), (x_{2} = b, y_{2} = c) and (x_{3} = c, y_{3} = a)

Let the centroid be (x, y)

Then,

\(x=\frac{1}{3}(x_{1}+x_{2}+x_{3})\)

\(x=\frac{1}{3}(a +b + c)\)

= \(\frac{a+b+c}{3}\)

\(y =\frac{1}{3}(y_{1}+ y_{2}+ y_{3})\)

\(y =\frac{1}{3}(b + c + a)\)

= \(\frac{a+b+c}{3}\)

But it is given that the centroid of the triangle is the origin

Then, we have:

\(\frac{a+b+c}{3}=0\)

\(\Rightarrow a+b+c=0\)

**Question 15:**

**Find the centroid of triangle ABC whose vertices are A(2,2),** **B (-4, -4) and C (5, -8)**

The given points are A (2, 2), B (-4, -4) and C (5, -8)

Here,

(x_{1} = 2, y_{1} = 2), (x_{2} = -4, y_{2} = -4) and (x_{3} = 5, y_{3} = -8)

Let G (x, y) be the centroid of \(\Delta ABC\).Then,

\(x=\frac{1}{3}(x_{1}+x_{2}+x_{3})\)

=\(\frac{1}{3}(2 – 4 + 5)\)

= 1

\(y=\frac{1}{3}(y_{1}+ y_{2}+ y_{3})\)

= \(\frac{1}{3}(2 – 4 – 8)\)

= \(\frac{-10}{3}\)

Hence, the centroid of \(\Delta ABC\) is G\(\left (1,\frac{-10}{3}\right )\)

**Question 16:**

**In what ratio does the point C(4,5) divide the join of A(2, 3) and B(7, 8) ?**

Let the required ratio be k: 1

Then, by section formula, the co-ordinates of C are

C\(\left (\frac{7k+2}{k+1},\frac{8k+3}{k+1} \right )\)

Therefore,

\(\frac{7k+2}{k+1}=4\;and\;\frac{8k+3}{k+1}=5\)\(\left [Because, C(4,5)\;is\;given\right ]\)

\(\Rightarrow \) 7k + 2 = 4k + 4 and 8k + 3 = 5k +5

\(\Rightarrow \) 3k = 2 and 3k = 2

\(\Rightarrow k=\frac{2}{3}\) in each case

So, the required ratio is \(\frac{2}{3}:1\), which is same as 2:3

**Question 17:**

**If the points A(2, 3) , B(4, k) and C(6, -3) are collinear, find the value of k.**

The given points are A(2, 3) , B(4, k) and C(6, -3)

Here

(x_{1} = 2 , y_{1} = 3), (x_{2} = 4 , y_{2} = k) and (x_{3} = 6 , y_{3} = -3)

It is given that the points A, B and C are collinear. Then

x_{1}(y_{2} â€“ y_{3}) + x_{2}(y_{3} â€“ y_{1}) + x_{3}(y_{1} â€“ y_{2}) = 0

\(\Rightarrow \) 2(k + 3) + 4(-3 â€“ 3) +6(3 â€“ k) = 0

\(\Rightarrow \) 2k + 6 -24 +18 -6k = 0

\(\Rightarrow \) -4k = 0

\(\Rightarrow \) k = 0′