Coordinate GeometryClass 10 Mathematics NCERT Solutions

Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given: Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (4, 1)$.

Solution: Distance $= \sqrt{(4 - 2)^2 + (1 - 3)^2}$ $= \sqrt{(2)^2 + (-2)^2}$ $= \sqrt{4 + 4}$ $= \sqrt{8}$ $= 2\sqrt{2}$

Final Answer: The distance between the points $(2,3)$ and $(4,1)$ is $2\sqrt{2}$ units.

Given: Let $(x_1, y_1) = (-5, 7)$ and $(x_2, y_2) = (-1, 3)$.

Solution: Distance $= \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$ $= \sqrt{(-1 + 5)^2 + (-4)^2}$ $= \sqrt{(4)^2 + (-4)^2}$ $= \sqrt{16 + 16}$ $= \sqrt{32}$ $= 4\sqrt{2}$

Final Answer: The distance between the points $(-5,7)$ and $(-1,3)$ is $4\sqrt{2}$ units.

Given: Let $(x_1, y_1) = (a, b)$ and $(x_2, y_2) = (-a, -b)$.

Solution: Distance $= \sqrt{(-a - a)^2 + (-b - b)^2}$ $= \sqrt{(-2a)^2 + (-2b)^2}$ $= \sqrt{4a^2 + 4b^2}$ $= \sqrt{4(a^2 + b^2)}$ $= 2\sqrt{a^2 + b^2}$

Final Answer: The distance between the points $(a, b)$ and $(-a, -b)$ is $2\sqrt{a^2 + b^2}$ units.

Given: Let the points be $O(0,0)$ and $P(36,15)$. So, $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (36, 15)$.

Formula: The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For distance from the origin, this simplifies to $d = \sqrt{x^2 + y^2}$.

Solution: Distance OP $= \sqrt{(36 - 0)^2 + (15 - 0)^2}$ $= \sqrt{36^2 + 15^2}$ $= \sqrt{1296 + 225}$ $= \sqrt{1521}$ $= 39$

Final Answer (Part 1): The distance between the points $(0,0)$ and $(36,15)$ is 39 units.

Given: The problem in Section 7.2 describes town B located 36 km east and 15 km north of town A. If we consider town A to be at the origin $(0,0)$ of a coordinate system, then the coordinates of town B would be $(36, 15)$.

Solution: The distance between town A and town B is the distance between the points $(0,0)$ and $(36,15)$, which we calculated in Part 1.

Final Answer (Part 2): Yes, we can find the distance. The distance between the two towns A and B is 39 km.

Given: Let the points be $A(1,5)$, $B(2,3)$, and $C(-2,-11)$.

Condition for Collinearity: Three points are collinear if the sum of the lengths of any two line segments among them is equal to the length of the remaining line segment. For example, if $AB + BC = AC$, the points are collinear.

Formula: The distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Solution: We calculate the distances between each pair of points.

Distance AB: $AB = \sqrt{(2 - 1)^2 + (3 - 5)^2}$ $= \sqrt{1^2 + (-2)^2}$ $= \sqrt{1 + 4} = \sqrt{5}$

Distance BC: $BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2}$ $= \sqrt{(-4)^2 + (-14)^2}$ $= \sqrt{16 + 196} = \sqrt{212}$

Distance AC: $AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2}$ $= \sqrt{(-3)^2 + (-16)^2}$ $= \sqrt{9 + 256} = \sqrt{265}$

Now, we check if the sum of any two distances equals the third. $AB + BC = \sqrt{5} + \sqrt{212}$ $AB + AC = \sqrt{5} + \sqrt{265}$ $BC + AC = \sqrt{212} + \sqrt{265}$

Clearly, $AB + BC \neq AC$. Also, $AB + AC \neq BC$ and $BC + AC \neq AB$.

Since the sum of the lengths of any two segments is not equal to the length of the third segment, the points do not lie on the same line.

Final Answer: The points $(1,5), (2,3)$ and $(-2,-11)$ are not collinear.

Given: Let the vertices of the triangle be $A(5,-2)$, $B(6,4)$, and $C(7,-2)$.

Condition for an Isosceles Triangle: A triangle is isosceles if at least two of its sides have equal length.

Formula: The distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Solution: We calculate the lengths of the three sides of the triangle ABC.

Length of side AB: $AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2}$ $= \sqrt{1^2 + (4 + 2)^2}$ $= \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$

Length of side BC: $BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2}$ $= \sqrt{1^2 + (-6)^2}$ $= \sqrt{1 + 36} = \sqrt{37}$

Length of side AC: $AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2}$ $= \sqrt{2^2 + (-2 + 2)^2}$ $= \sqrt{2^2 + 0^2} = \sqrt{4} = 2$

Comparing the side lengths, we see that $AB = BC = \sqrt{37}$.

Since two sides of the triangle are equal, the triangle is an isosceles triangle.

Final Answer: Yes, the points $(5,-2)$, $(6,4)$, and $(7,-2)$ are the vertices of an isosceles triangle.

Given: From the description of the seating arrangement, which can be represented on a coordinate plane, the positions of the four friends are: $A(3,4)$, $B(6,7)$, $C(9,4)$, and $D(6,1)$.

To Determine: Whether the quadrilateral ABCD formed by these points is a square.

Formula: The distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Solution: We calculate the lengths of the four sides and the two diagonals.

Side lengths: $AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. $BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. $CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. $DA = \sqrt{(3-6)^2 + (4-1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.

Since $AB = BC = CD = DA = 3\sqrt{2}$, all four sides are equal.

Diagonal lengths: $AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6$ units. $BD = \sqrt{(6-6)^2 + (1-7)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6$ units.

Since $AC = BD = 6$, the diagonals are also equal.

Because all four sides are equal and both diagonals are equal, the quadrilateral ABCD is a square.

Conclusion: Champa's observation that ABCD is a square is correct. Chameli disagrees, so Chameli is incorrect.

Final Answer: Champa is correct.

Solution: We find the lengths of all four sides and the two diagonals. $AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ $BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ $CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ $DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ All sides are equal: $AB = BC = CD = DA$.

Now, we check the diagonals: $AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ $BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ The diagonals are also equal: $AC = BD$.

Reason: Since all four sides are equal and the diagonals are equal, the quadrilateral is a square.

Final Answer (i): Square.

Solution: We find the lengths of the sides. $AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$ $BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$ $CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$ $DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$ No sides are equal. Let us check for collinear points. For points B, C, D: $BC = \sqrt{13}$, $CD = \sqrt{50}$, $BD = \sqrt{(-1-3)^2 + (-4-1)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41}$. $BC + BD \neq CD$. So, no three points are collinear.

Reason: The points form a quadrilateral with no special properties like equal sides or parallel sides (slopes are all different). It is a general quadrilateral.

Final Answer (ii): A general quadrilateral.

Solution: We find the lengths of all four sides and the two diagonals. $AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ $BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ $CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$ $DA = \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ Opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = \sqrt{18}$.

Now, we check the diagonals: $AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$ $BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$ The diagonals are not equal: $AC \neq BD$.

Reason: Since opposite sides are equal but the diagonals are not equal, the quadrilateral is a parallelogram.

Final Answer (iii): Parallelogram.

Given: Let the point on the x-axis be $P(x, 0)$. Let the given points be $A(2,-5)$ and $B(-2,9)$. The point P is equidistant from A and B, which means $PA = PB$.

Condition: $PA = PB \implies PA^2 = PB^2$

Formula: The square of the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2$.

Solution: $PA^2 = (x - 2)^2 + (0 - (-5))^2 = (x - 2)^2 + 5^2$ $PB^2 = (x - (-2))^2 + (0 - 9)^2 = (x + 2)^2 + (-9)^2$

Setting $PA^2 = PB^2$: $(x - 2)^2 + 25 = (x + 2)^2 + 81$

Expanding the terms: $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$ $x^2 - 4x + 29 = x^2 + 4x + 85$

Subtract $x^2$ from both sides: $-4x + 29 = 4x + 85$

Rearranging the terms to solve for $x$: $29 - 85 = 4x + 4x$ $-56 = 8x$ $x = \frac{-56}{8}$ $x = -7$

The point on the x-axis is $P(-7, 0)$.

Final Answer: The required point on the x-axis is $(-7, 0)$.

Given: The points are $P(2,-3)$ and $Q(10,y)$. The distance between P and Q is 10 units, i.e., $PQ = 10$.

Formula: The distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. So, $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

Solution: We are given $PQ = 10$, so $PQ^2 = 10^2 = 100$.

Using the distance formula: $PQ^2 = (10 - 2)^2 + (y - (-3))^2$ $100 = (8)^2 + (y + 3)^2$ $100 = 64 + (y + 3)^2$

Now, solve for $y$: $100 - 64 = (y + 3)^2$ $36 = (y + 3)^2$

Taking the square root of both sides: $\sqrt{36} = y + 3$ $\pm 6 = y + 3$

This gives two possible values for $y$: Case 1: $6 = y + 3 \implies y = 6 - 3 = 3$ Case 2: $-6 = y + 3 \implies y = -6 - 3 = -9$

Final Answer: The possible values of $y$ are $3$ and $-9$.

Given: The point $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$. This means $QP = QR$, which implies $QP^2 = QR^2$.

Formula: The square of the distance formula: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

Solution: $QP^2 = (5 - 0)^2 + (-3 - 1)^2 = 5^2 + (-4)^2 = 25 + 16 = 41$. $QR^2 = (x - 0)^2 + (6 - 1)^2 = x^2 + 5^2 = x^2 + 25$.

Setting $QP^2 = QR^2$: $41 = x^2 + 25$ $x^2 = 41 - 25$ $x^2 = 16$ $x = \pm \sqrt{16}$ $x = 4$ or $x = -4$.

Final Answer (Part 1): The values of $x$ are $4$ and $-4$.

We need to find the distances for both values of $x$.

Case 1: $x = 4$ The point R is $(4,6)$.

Distance QR: $QR = \sqrt{(4 - 0)^2 + (6 - 1)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$ units.

Distance PR: $PR = \sqrt{(4 - 5)^2 + (6 - (-3))^2} = \sqrt{(-1)^2 + (6+3)^2} = \sqrt{1 + 9^2} = \sqrt{1 + 81} = \sqrt{82}$ units.

Case 2: $x = -4$ The point R is $(-4,6)$.

Distance QR: $QR = \sqrt{(-4 - 0)^2 + (6 - 1)^2} = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$ units.

Distance PR: $PR = \sqrt{(-4 - 5)^2 + (6 - (-3))^2} = \sqrt{(-9)^2 + (6+3)^2} = \sqrt{81 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$ units.

Final Answer (Part 2): When $x=4$, $QR = \sqrt{41}$ and $PR = \sqrt{82}$. When $x=-4$, $QR = \sqrt{41}$ and $PR = 9\sqrt{2}$.

Given: Let the point be $P(x,y)$. Let the given points be $A(3,6)$ and $B(-3,4)$. The point P is equidistant from A and B, which means $PA = PB$.

Condition: $PA = PB \implies PA^2 = PB^2$

Formula: The square of the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2$.

Solution: $PA^2 = (x - 3)^2 + (y - 6)^2$ $PB^2 = (x - (-3))^2 + (y - 4)^2 = (x + 3)^2 + (y - 4)^2$

Setting $PA^2 = PB^2$: $(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$

Expanding the terms: $(x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16)$

Combine like terms on each side: $x^2 + y^2 - 6x - 12y + 45 = x^2 + y^2 + 6x - 8y + 25$

Cancel $x^2$ and $y^2$ from both sides: $-6x - 12y + 45 = 6x - 8y + 25$

Move all terms to one side to form the relation: $45 - 25 = 6x + 6x - 8y + 12y$ $20 = 12x + 4y$

Divide the entire equation by 4 to simplify: $5 = 3x + y$

Or, written in standard form: $3x + y - 5 = 0$

Final Answer: The required relation between $x$ and $y$ is $3x + y - 5 = 0$.

Given: The line segment joins the points $A(-1,7)$ and $B(4,-3)$. The ratio of division is $m_1:m_2 = 2:3$. Let the point of division be $P(x,y)$. Here, $(x_1, y_1) = (-1, 7)$ and $(x_2, y_2) = (4, -3)$.

Formula: The section formula for a point $P(x,y)$ that divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m_1:m_2$ is: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$ $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Solution: Calculating the x-coordinate: $x = \frac{2(4) + 3(-1)}{2 + 3}$ $x = \frac{8 - 3}{5}$ $x = \frac{5}{5} = 1$

Calculating the y-coordinate: $y = \frac{2(-3) + 3(7)}{2 + 3}$ $y = \frac{-6 + 21}{5}$ $y = \frac{15}{5} = 3$

The coordinates of the point are $(1,3)$.

Final Answer: The coordinates of the required point are $(1,3)$.

Given: Let the line segment be AB, with $A(4,-1)$ and $B(-2,-3)$. The points of trisection are two points, let's call them P and Q, that divide the segment AB into three equal parts, i.e., $AP = PQ = QB$.

Finding point P: Point P divides the segment AB in the ratio $1:2$. Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, and $m_1:m_2 = 1:2$.

Formula: Using the section formula: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$, $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Solution for P: $x_P = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$ $y_P = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 - 2}{3} = \frac{-5}{3}$ So, the coordinates of point P are $(2, -\frac{5}{3})$.

Finding point Q: Point Q divides the segment AB in the ratio $2:1$. (Or, Q is the midpoint of PB). Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, and $m_1:m_2 = 2:1$.

Solution for Q: $x_Q = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = \frac{0}{3} = 0$ $y_Q = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3}$ So, the coordinates of point Q are $(0, -\frac{7}{3})$.

Final Answer: The coordinates of the points of trisection are $(2, -\frac{5}{3})$ and $(0, -\frac{7}{3})$.

Setting up the Coordinates: Let's consider the ground as a coordinate plane where the lines are parallel to the y-axis and AD is along the y-axis. The distance along AD is 100 m (since there are 100 pots at 1m distance).

Position of Niharika's Green Flag (G): Niharika is on the 2nd line, so her x-coordinate is 2. She runs $\frac{1}{4}$th the distance AD, so her y-coordinate is $\frac{1}{4} \times 100 = 25$. Coordinates of the green flag G are $(2, 25)$.

Position of Preet's Red Flag (R): Preet is on the 8th line, so her x-coordinate is 8. She runs $\frac{1}{5}$th the distance AD, so her y-coordinate is $\frac{1}{5} \times 100 = 20$. Coordinates of the red flag R are $(8, 20)$.

Formula: Distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Solution: Distance GR $= \sqrt{(8 - 2)^2 + (20 - 25)^2}$ $= \sqrt{6^2 + (-5)^2}$ $= \sqrt{36 + 25} = \sqrt{61}$

Final Answer (Part 1): The distance between both the flags is $\sqrt{61}$ m.

Rashmi posts a blue flag exactly halfway between G and R. This is the midpoint of the line segment GR.

Formula: Midpoint formula $M(x,y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.

Solution: Let the position of the blue flag be B(x,y). $x = \frac{2 + 8}{2} = \frac{10}{2} = 5$ $y = \frac{25 + 20}{2} = \frac{45}{2} = 22.5$

The coordinates for Rashmi's flag are $(5, 22.5)$. This means she should post her flag on the 5th line at a distance of 22.5 m from the starting point along that line.

Final Answer (Part 2): Rashmi should post her flag on the 5th line at a distance of 22.5 m.

Given: Let the line segment be AB, with $A(-3,10)$ and $B(6,-8)$. Let the point of division be $P(-1,6)$. Let the ratio in which P divides AB be $m_1:m_2$.

Formula: Section formula: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

Solution: Using the x-coordinate: $-1 = \frac{m_1(6) + m_2(-3)}{m_1 + m_2}$ $-1(m_1 + m_2) = 6m_1 - 3m_2$ $-m_1 - m_2 = 6m_1 - 3m_2$ $3m_2 - m_2 = 6m_1 + m_1$ $2m_2 = 7m_1$ $\frac{m_1}{m_2} = \frac{2}{7}$ So, the ratio $m_1:m_2 = 2:7$.

Verification using the y-coordinate: $y = \frac{2(-8) + 7(10)}{2 + 7} = \frac{-16 + 70}{9} = \frac{54}{9} = 6$. This matches the y-coordinate of P, so our ratio is correct.

Let the ratio be $k:1$. The coordinates of the dividing point are: $P(x,y) = (\frac{k(6) + 1(-3)}{k+1}, \frac{k(-8) + 1(10)}{k+1})$ We are given that $P(x,y) = (-1,6)$. Equating the x-coordinates: $-1 = \frac{6k - 3}{k+1}$ $-(k+1) = 6k - 3$ $-k - 1 = 6k - 3$ $2 = 7k$ $k = \frac{2}{7}$ The ratio is $k:1 = \frac{2}{7}:1$, which is $2:7$.

Final Answer: The ratio is $2:7$.

Given: The line segment joins $A(1,-5)$ and $B(-4,5)$. It is divided by the x-axis.

Understanding the problem: Any point on the x-axis has its y-coordinate equal to 0. So, let the point of division be $P(x,0)$. Let the ratio in which the x-axis divides the line segment AB be $k:1$.

Formula: Using the section formula, the coordinates of P are: $P(x,y) = (\frac{k x_2 + x_1}{k+1}, \frac{k y_2 + y_1}{k+1})$ Here, $(x_1, y_1) = (1,-5)$ and $(x_2, y_2) = (-4,5)$.

Solution: We use the y-coordinate of point P, which is 0. $y_P = \frac{k(5) + 1(-5)}{k+1} = 0$ $5k - 5 = 0$ $5k = 5$ $k = 1$

The ratio is $k:1 = 1:1$.

Final Answer (Part 1): The ratio is $1:1$.

Solution: Now that we know the ratio is $1:1$, we can find the x-coordinate of the point P. $x_P = \frac{1(-4) + 1(1)}{1+1}$ $x_P = \frac{-4 + 1}{2}$ $x_P = -\frac{3}{2}$

The point of division is $P(-\frac{3}{2}, 0)$. (Note: A ratio of 1:1 means the point of division is the midpoint.)

Final Answer (Part 2): The coordinates of the point of division are $(-\frac{3}{2}, 0)$.

Given: Let the vertices of the parallelogram be $A(1,2)$, $B(4,y)$, $C(x,6)$, and $D(3,5)$ taken in order.