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Correct ANSWER is (b) 26 units

Easiest EXPLANATION: Area of quadrilateral = Area of ∆ABC + Area of ∆ADC

We know that, area of triangle = \(\frac {1}{2}\){x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)}

The coordinates of VERTICES of the triangle are (8, 6), (9, 0) and (1, 2).

The area of triangle = \(\frac {1}{2}\){8(0 – 2) + 9(2 – 6) + 1(6 – 0)} = \(\frac {1}{2}\) { – 16 – 36 + 6} = \(\frac {-46}{2}\) = – 23 units

The area of triangle cannot be zero. So, Area of ∆ABC = 23 units

We know that, area of triangle = \(\frac {1}{2}\){x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)}

The coordinates of vertices of the triangle are (8, 6), (3, 4) and (1, 2).

The area of triangle = \(\frac {1}{2}\) {8(4 – 2) + 3(2 – 6) + 1(6 – 4)} = \(\frac {1}{2}\) {16 – 12 + 2} = \(\frac {6}{2}\) = 3 units

So, Area of ∆ADC = 3 units

Area of quadrilateral = Area of ∆ABC + Area of ∆ADC = 23 + 3 = 26 units

Right option is (b) 11 + √91, 11 – √91

The EXPLANATION is: Distance between (5, 11) and (2, x) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2-5)^2 + (x-11)^2} \)

= \( \sqrt {x^2-22X + 121 + (-3)^2} \)

= \( \sqrt {x^2-22x + 121 + 9} \)

= \( \sqrt {x^2-22x + 130} \)

The distance between (5, 11) and (2, x) is 10

∴ \( \sqrt {x^2-22x + 130} \) = 10

Squaring on both sides we get,

x^2 – 22x + 130 = 100

x^2 – 22x + 130 – 100 = 0

x^2 – 22x + 30 = 0

x = 11 + √91, 11 – √91

Correct answer is (a) a = 2, b = 3

To explain I would say: Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The POINTS are A(5, 8)and B(a, b)in the ratio 2:5

∴ x = \(\frac {2(a)+5(5)}{2+5} = \frac {2a+25}{7}\)

y = \(\frac {2(b)+5(8)}{2+5} = \frac {2b+40}{7}\)

But the COORDINATES of C are \((\frac {29}{7}, \frac {46}{7} )\)

Therefore, \(\frac {2a+25}{7} = \frac {29}{7}\)

a = 2

\(\frac {2b+40}{7} = \frac {46}{7}\)

b = 3

Right answer is (d) 2:1 (internally)

For EXPLANATION I WOULD say: Let the ratio in which the point (\(\frac {-19}{3}, \frac {7}{3}\)) divides the LINE segment joining the POINTS A(3, 7) and B(-11, 0) be k:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(3, 7) and B(-11, 0) and the ratio is k:1

∴ x = \(\frac {k(-11)+1(3)}{k+1} = \frac {-11k+3}{k+1}\)

y = \(\frac {k(0)+1(7)}{k+1} = \frac {7}{k+1}\)

Since, the point is (\(\frac {-19}{3}, \frac {7}{3}\)).

∴ \(\frac {-19}{3} = \frac {-11k+3}{k+1}\)

-19(k + 1) = 3(-11k + 3)

-19k – 19 = -33k + 9

-19k + 33k = 19 + 9

14k = 28

k = \(\frac {28}{14}\) = 2

The ratio is 2:1.

The correct option is (a) False

For EXPLANATION: We know that, area of triangle = \(\frac {1}{2}\){x1(Y2 – y3 ) + x2(y3 – y1 ) + x3(y1 – y2 )}

The three points are A (a, 0), B (0, b) and C (c, 0)

Since the points are collinear, the area of triangle will be zero.

The area of triangle = \(\frac {1}{2}\) {a(b – 0) + 0(c – 0) + c(0 – b) } = \(\frac {1}{2}\) {ab + 0 – bc}

\(\frac {1}{2}\) {ab – bc} = 0

{ab – bc} = 0

a = c

Correct answer is (d) a = 0, B = 1

The best EXPLANATION: PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram BISECT each other.

∴ O is the mid-point SQ and PR.

Midpoint of PR

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are P(-5, a) and R(-b, 0) and the ratio is 1:1

∴ x = \(\frac {1(-b)+1(-5)}{2} = \frac {-b-5}{2}\)

y = \(\frac {1(0)+1(a)}{2} = \frac {a}{2}\)

Midpoint of QS

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are Q(-3, -3) and S(-3, 3) and the ratio is 1:1

∴ x = \(\frac {1(-3)+1(-3)}{2} = \frac {-6}{2}\) = -3

y = \(\frac {1(3)+1(-3)}{2} = \frac {0}{2}\) = 0

Therefore, \(\frac {-b-5}{2}\) = -3

b = 1

\(\frac {a}{2}\) = 0

a = 0

Right OPTION is (d) (0, \( \FRAC {89}{18} \))

To EXPLAIN I would say: LET the point on y-axis be (0, y)

Distance between (-1, 0) and (0, y) = \( \SQRT {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0 + 1)^2 + (y-0)^2} \)

= \( \sqrt {y^2 + (1)^2} \)

= \( \sqrt {y^2 + 1} \)

Distance between (3, 9) and (0, y) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0-3)^2 + (y-9)^2} \)

= \( \sqrt {y^2-18y + 81 + (-3)^2} \)

= \( \sqrt {y^2-18y + 81 + 9} \)

= \( \sqrt {y^2-18y + 90} \)

Since, the point (0, y) is equidistant from (-1, 0) and (3, 9)

The distances will be equal

∴ \( \sqrt {y^2 + 1} =\sqrt {y^2-18y + 90} \)

Squaring on both sides we get,

y^2 + 1 = y^2 – 18y + 90

1 – 90 = -18y

-89 = -18y

y = \( \frac {89}{18} \)

The point is (0, \( \frac {89}{18} \))

Correct choice is (c) -15 UNITS

To explain I would say: We know that, area of triangle = \(\FRAC {1}{2}\){x1(y2 – Y3) + x2(y3 – y1) + X3(y1 – y2 )}

The coordinates of vertices of the triangle are (3, 1), (0, 4) and (5, 9).

The area of triangle = \(\frac {1}{2}\){3(4 – 9) + 0(9 – 1) + 5(1 – 4) } = \(\frac {1}{2}\) {-15 + 0 – 15} = \(\frac {-30}{2}\) = -15 units

Since area of triangle cannot be zero. So area of triangle is 15 units.

Right choice is (c) 2:9 (EXTERNALLY)

Easy explanation: Let the ratio in which the x-axis divides the LINE SEGMENT JOINING the points A(-5, 2) and B(3, 9) be k:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-5, 2) and B(3, 9) and the ratio is k:1

∴ x = \(\frac {k(3)+1(-5)}{k+1} = \frac {3k-5}{k+1}\)

y = \(\frac {k(9)+1(2)}{k+1} = \frac {9k+2}{k+1}\)

Since, the point is on x-axis.

Hence, the y-coordinate will be zero.

∴ 0 = \(\frac {9k+2}{k+1}\)

0 = 9k+2

k = \(\frac {-2}{9}\)

The ratio in which the y-axis cuts the line segment joining the points A(-5, 2) and B(3, 9) will be 2:9 (externally).

Right choice is (b) (\(\frac {-4}{3}, \frac {20}{9}\))

Best explanation: Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-2, 2) and B(-1, 5) and the RATIO is 2:5

∴ x = \(\frac {2(-1)+5(-2)}{2+7} = \frac {-2-10}{9} = \frac {-12}{9} = \frac {-4}{3}\)

y = \(\frac {2(5)+5(2)}{2+7} = \frac {10+10}{9} = \frac {20}{9} = \frac {20}{9}\)

HENCE, the point is (\(\frac {-4}{3}, \frac {20}{9}\)).

The CORRECT answer is (d) 3:1 (externally)

Explanation: Let the ratio in which the y-axis divides the line segment joining the points A(2, 4) and B(6, 5) be K:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(2, 4) and B(6, 5) and the ratio is k:1

∴ x = \(\frac {k(6)+1(2)}{k+1} = \frac {6K+2}{k+1}\)

y = \(\frac {k(5)+1(4)}{k+1} = \frac {5k+4}{k+1}\)

Since, the point is on y-axis.

Hence, the x-coordinate will be zero.

∴ 0 = \(\frac {6k+2}{k+1}\)

0 = 6k + 2

k = \(\frac {-6}{2}\) = -3

The ratio in which the y-axis CUTS the line segment joining the points A(2, 4) and B(6, 5) will be 3:1 (externally).

Right answer is (d) (5, -2)

The best I can EXPLAIN: The two vertices of TRIANGLE are A (-3, 0) and B (-8, 5). Its centroid is (-2, 1).

We KNOW, xcentroid = \(\frac {x_1+x_2+x_3}{3}\) and ycentroid = \(\frac {y_1+y_2+y_3}{3}\)

Now, xcentroid = \(\frac {-3-8+x_3}{3}\)

xcentroid = -2

-2 = \(\frac {-3 – 8 + x_3}{3}\)

-6 = -3 – 8 + x3

5 = x3

Now, ycentroid = \(\frac {0+5+y_3}{3}\)

ycentroid = 1

1 = \(\frac {0 + 5 + y_3}{3}\)

3 = 5 + y3

-2 = y3

The THIRD coordinate is (5, -2).

The correct option is (b) False

Best explanation: We KNOW that, area of triangle = \(\FRAC {1}{2}\){x1 (y2 – Y3 ) + X2 (y3 – y1 ) + x3 (y1 – y2 )}

Let us assume the vertices of the triangle are A (3, 0), B (4, 5) and C (6, 7)

The area of triangle = \(\frac {1}{2}\) {3(5 – 7) + 4(7 – 0) + 6(0 – 5)} = \(\frac {1}{2}\) {-8} = – 4 units

Since, the area of triangle is not zero.

Therefore, the points are not collinear.

The correct option is (a) (\(\FRAC {2}{3}\), 2)

Explanation: We know, xcentroid = \(\frac {x_1+x_2+x_3}{3}\) and ycentroid = \(\frac {y_1+y_2+y_3}{3}\)

xcentroid = \(\frac {-2+0+4}{3} = \frac {2}{3}\)

ycentroid = \(\frac {4+0+2}{3}\) = 2

The coordinates of the centroid are (\(\frac {2}{3}\), 2).

Correct answer is (c) (5, 6)

To explain I would say: Midpoint lies in the center of the LINE segment

Hence, it divides the line in the RATIO 1:1

Using, section formula X = \(\FRAC {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-5, 10) and B(15, 2) and the ratio is 1:1

∴ x = \(\frac {1(-5)+1(15)}{2} = \frac {-5+15}{2} = \frac {10}{2}\) = 5

y = \(\frac {1(2)+1(10)}{2} = \frac {2+10}{2} = \frac {12}{2}\) = 6

Hence, the point is (5,6).

Correct choice is (b) \(\frac {3}{2}\) units

Easiest explanation: ABC is the triangle formed by the coordinates A (3, 3), B (9, 9) and C (4, 6) and DEF is the triangle formed by joining the MIDPOINTS of AC, BC, AB.

D is the midpoint of AC. Coordinates of D = \( ( \frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {3+4}{2}, \frac {3+6}{2} ) = ( \frac {7}{2}, \frac {9}{2} ) \)

E is the midpoint of AB. Coordinates of E = \( ( \frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {3+9}{2}, \frac {3+9}{2} ) = ( \frac {12}{2}, \frac {12}{2} ) \) = (6, 6)

F is the midpoint of BC. Coordinates of F = \( ( \frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {9+4}{2}, \frac {9+6}{2} ) = ( \frac {13}{2}, \frac {15}{2} ) \)

The AREA of triangle formed by D \( ( \frac {7}{2}, \frac {9}{2} ) \), E (6, 6) and F\( ( \frac {13}{2}, \frac {15}{2} ) \)

We know that, area of triangle = \(\frac {1}{2}\){x1 (y2 – y3 ) + X2 (y3 – y1 ) + x3 (y1 – y2 )}

The coordinates of vertices of the triangle are \( ( \frac {7}{2}, \frac {9}{2} ) \), (6, 6) and \( ( \frac {13}{2}, \frac {15}{2} ) \).

The area of triangle = \(\frac {1}{2} \{ \frac {7}{2}\) (6 – \(\frac {15}{2}\)) + 6(\(\frac {15}{2} – \frac {9}{2}\)) + \(\frac {13}{2} ( \frac {9}{2} – 6 ) \} \) = \(\frac {1}{2} \{ – \frac {21}{4}\) + 18 – \(\frac {39}{4} \} = \frac {3}{2} \) units

The correct answer is (a) √153

The EXPLANATION: Using distance formula,

Distance between (5, 7) and (8, -5) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(8-5)^2 + (-5-7)^2}\)

= \( \sqrt {(3)^2 + (-12)^2} \)

= \( \sqrt {9 + 144}\)

= √153

The correct CHOICE is (c) (5, 12)

To explain: PQRS is a parallelogram. The OPPOSITE side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.

∴ O is the mid-point SQ and PR.

MIDPOINT of PR

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are P(-1, -1) and R(2, 3) and the RATIO is 1:1

∴ x = \(\frac {1(-1)+1(2)}{2} = \frac {-1+2}{2} = \frac {1}{2}\)

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

Hence, the coordinates of O is (5, 6)

Midpoint of QS

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are Q(2, 0) and S(a, b) and the ratio is 1:1

∴ x = \(\frac {1(a)+1(2)}{2} = \frac {a+2}{2}\)

y = \(\frac {1(b)+1(0)}{2} = \frac {b}{2}\)

The coordinates of O is (5, 6)

Therefore, \(\frac {a+2}{2}\) = 5

a = 8

\(\frac {b}{2}\) = 6, b = 12

The coordinates of S are (5, 12).

Right choice is (b) False

The best I can explain: We know that, area of TRIANGLE = \(\frac {1}{2}\){X1 (y2 – y3 ) + x2 (y3 – Y1 ) + x3 (y1 – y2 )}

LET us assume the vertices of the triangle are (0, 0), (0, 4) and (0, 9).

The area of triangle = \(\frac {1}{2}\) {0(4 – 9) + 0(9 – 0) + 0(0 – 4) } = \(\frac {1}{2}\){0} = 0 units

Hence, the area of triangle will be zero if the points are collinear.

Right option is (a) y = \(\frac {9}{2}\)

The explanation: Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are (-1, 4)and B(6, 5)in the ratio 1:3

∴ x = \(\frac {1(6)+3(-1)}{1+3} = \frac {6-3}{4} = \frac {3}{4}\)

y = \(\frac {1(6)+3(4)}{1+3} = \frac {6+12}{4} = \frac {18}{4}\)

Therefore y = \(\frac {9}{2}\)

Right CHOICE is (c) 6 units

To EXPLAIN: Distance between A (-1, -1) and B (-1, 3) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-1 + 1)^2 + (3 + 1)^2} \)

= \( \sqrt {4^2} \)

= √16

= 4

Distance between B (-1, 3) and C(2, -1) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2 + 1)^2 + (-1-3)^2} \)

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

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

= 5

Distance between A (-1, -1) and C (2, -1) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2 + 1)^2 + (-1 + 1)^2} \)

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

= 3

Now, AC^2 + AB^2 = 4^2 + 3^2 = 16 + 9 = 25

BC^2 = 5^2 = 25

Hence, it is a right-angled triangle, right-angled at A.

Area of triangle = \( \FRAC {1}{2}\) × base × height = \( \frac {1}{2}\) × 4 × 3 = 6 units

The correct answer is (d) (4 + \( \SQRT {11I} \), 0), (4 – \( \sqrt {11i} \), 0)

For explanation I would say: Let the POINT on x-axis be (x, 0)

Distance between (4, 6) and (x, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(x-4)^2 + (0-6)^2} \)

= \( \sqrt {x^2-8x + 16 + 36} \)

= \( \sqrt {x^2-8x + 52} \)

The distance between (4, 6) and (x, 0) is 12

∴ \( \sqrt {x^2-8x + 52} \) = 12

Squaring on both sides, we get,

x^2 – 8x + 52 = 25

x^2 – 8x + 27 = 0

x = 4 + \( \sqrt {11i} \), 4 – \( \sqrt {11i} \)

The correct answer is (b) True

Easiest explanation: Consider three points (-3, 3), (-1, 2) and (1, 1)

We know that, area of triangle = \(\frac {1}{2}\){x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2)}

The area of triangle = \(\frac {1}{2}\) {-3(2 – 1) + (-1)(1 – 3) + 1(3 – 2)} = \(\frac {1}{2}\) {-3 – 1 + 3 + 3 – 2} = \(\frac {0}{2}\) = 0

Hence if the points are collinear the area of triangle is zero.

Correct answer is (a) True

Explanation: A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC.

Draw AL, CM, BN perpendicular to x – axis.

Then, ML = (x1 – x2), LN = (x3 – x1) and MN = (x3 – x2).

∴ AREA of ∆ABC = AR(trap.BMLA) + ar(trap.ALNC) + ar(trap.BMNC)

 = {\(\frac {1}{2}\) (AL + BM) × ML} + {\(\frac {1}{2}\) (AL + CN) × LN} – {\(\frac {1}{2}\) (CN + BM) × MN}

 = {\(\frac {1}{2}\) (y1 + y2 ) × (x1 – x2)} + {\(\frac {1}{2}\) (y1 + y3 ) × (x3 – x1)} – {\(\frac {1}{2}\) (y2 + y3 ) × (x3 – x2)}

 = \(\frac {1}{2}\) {x1 (y1 + y2 – y1 – y3 ) + x2 (y2 + y3 – y1 – y2 ) – x3 (y1 + y3 – y2 – y3 ) }

 = \(\frac {1}{2}\) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }

Correct choice is (b) 1:2 (externally)

The EXPLANATION is: Let the RATIO in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4) be k:1.

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(0, -1) and B(-3, -4) and the ratio is k:1.

∴ x = \(\frac {k(-3)+1(0)}{k+1} = \frac {-3k}{k+1}\)

y = \(\frac {k(-4)+1(-1)}{k+1} = \frac {-4k-1}{k+1}\)

Since, the point \((\frac {-3k}{k+1}, \frac {-4k-1}{k+1} )\) lies on the line 3x+y-11 = 0.

3 \((\frac {-3k}{k+1} + \frac {-4k-1}{k+1} )\)-11 = 0

3(-3k) + (-4k – 1) – 11(k + 1) = 0

-9k – 4k – 1 – 11K – 11 = 0

-24k – 12 = 0

-24k = 12

k = \(\frac {12}{-24} = \frac {-1}{2}\)

The ratio is 1:2 (externally).

Right answer is (B) Isosceles

The explanation is: Distance between (0, 3) and (5, 0) = \( \SQRT {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(5-0)^2 + (0-3)^2} \)

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

= \( \sqrt {25 + 9} \)

= √34

Distance between (5, 0) and (-5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-5-5)^2 + (0-0)^2} \)

= \( \sqrt {-10^2} \)

= 10

Distance between (0, 3) and (-5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-5-0)^2 + (0-3)^2} \)

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

= \( \sqrt {25 + 9} \)

= √34

Since, the TWO sides of the triangle are equal.

Hence, the triangle will be isosceles triangle.

Right answer is (a) a + b = -3

Easy explanation: The point is (a, b)

Distance between (3, 1) and (a, b) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(a-3)^2 + (b-1)^2} \)

= \( \sqrt {a^2-6a + 9 + b^2-2b + 1} \)

= \( \sqrt {a^2-6a + 10 + b^2-2b} \)

Distance between (2, 0) and (a, b) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(a-2)^2 + (b-0)^2} \)

= \( \sqrt {a^2-4a + 4 + b^2 } \)

Since, the point (a, b) is equidistant from (-1, 0) and (3, 9)

The distances will be equal

∴ \( \sqrt {a^2-6a + 10 + b^2-2b} = \sqrt {a^2-4a + 4 + b^2 } \)

SQUARING on both SIDES we get,

a^2 – 6a + 10 + b^2 – 2b = a^2 – 4a + 4 + b^2

-6a + 10 – 2b = -4a + 4

-2a – 6 = 2b

-a – b = 3

a + b = -3

Right choice is (b) (-2, 3)

For explanation I would say: We know that the diameter is twice the radius.

Hence, the CENTER is the midpoint of the diameter.

Using, section FORMULA x = \(\FRAC {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-2, -3) and center is (-2, 0) and the ratio is 1:1

Let the coordinates of other side of the radius be (x, y).

∴ -2 = \(\frac {1(-2)+1(x)}{2} = \frac {-2+x}{2}\)

-4 = -2 + x

-4 + 2 = x

x = -2

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

0 = -3 + y

y = 3

Hence, the point is (-2, 3).