Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. | |
The coordinates \((x, y)\) of the point that divides the line segment joining \((-1, 7)\) and \((4, -3)\) in the ratio \(2:3\) can be found using the section formula. The section formula is given by: \[ x = \frac{m_1 \cdot x_2 + m_2 \cdot x_1}{m_1 + m_2} \] \[ y = \frac{m_1 \cdot y_2 + m_2 \cdot y_1}{m_1 + m_2} \] In this case, \(m_1 : m_2 = 2:3\). Therefore, \(m_1 = 2\) and \(m_2 = 3\). Now, substitute the values into the formulas: \[ x = \frac{2 \cdot 4 + 3 \cdot (-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5}= 1 \] \[ y = \frac{2 \cdot (-3) + 3 \cdot 7}{2 + 3}= \frac{-6 + 21}{5} = \frac{15}{5}= 3 \] So, the coordinates of the point dividing the line segment in the ratio 2:3 are \((1, 3)\). |
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). | |
Assume that \( P (x_1, y_1) \) and \( Q (x_2, y_2) \) is the points of trisection of the line segment joining the given points \((4, -1)\) and \((-2, -3)\), so we can write \( AP = PQ = QB \). Therefore, point \( P \) divides \( AB \) in the ratio \( 1:2 \). By using section formula: \[ x_1 = \frac{1 \cdot (-2) + 2 \cdot 4}{3} = \frac{-2 + 8}{3} = \frac{6}{3} = 2 \] \[ y_1 = \frac{1 \cdot (-3) + 2 \cdot (-1)}{3} = \frac{-3 - 2}{3} = \frac{-5}{3} \] Therefore: \( P (x_1, y_1) = P(2, -\frac{5}{3}) \) Point \( Q \) divides \( AB \) in the ratio \( 2:1 \). By using section formula: \[ x_2 = \frac{2 \cdot (-2) + 1 \cdot 4}{3} = \frac{-4 + 4}{3} = 0 \] \[ y_2 = \frac{2 \cdot (-3) + 1 \cdot (-1)}{3} = \frac{-6 - 1}{3} = -\frac{7}{3} \] Therefore, the coordinates of the point \( Q \) are \( (0, -\frac{7}{3}) \). |
To conduct Sports Day activities, in your rectangular shaped school ground \( ABCD \), lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along \( AD \), as shown in Fig. 7.12. Niharika runs \(\frac{1}{4}\)th the distance \( AD \) on the 2nd line and posts a green flag. Preet runs \(\frac{1}{5}\)th the distance \( AD \) on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?  | |
As per question, Niharika posted the green flag at \( \frac{1}{4} \)th of the distance \( AD \), So, \( \left(\frac{1}{4} \times 100\right) \) m = 25 m from the starting point of the 2nd line. Therefore, We gets coordinates of this point are (2, 25). Similarly, Preet posted a red flag at \( \frac{1}{5} \) th of the distance \( AD \), So, \( \left(\frac{1}{5} \times 100\right) \) m = 20 m from the starting point of the 8th line. Therefore, We gets coordinates of this point are (8, 20). The distance between these flags: \[ \sqrt{{(8 - 2)^2 + (20 - 25)^2}} \] \[ \sqrt{{(6)^2 + ( - 5)^2}} =\sqrt{{36 + 25}}= \sqrt{{61}} \] So the distance between both flag is \(\sqrt{{61}} \) The point at which Rashmi should post her blue flag is the midpoint of the line joining these points. Assume that this point is \( P(x, y) \). \[ x = \frac{2 + 8}{2} = \frac{10}{2} = 5 \] \[ y = \frac{20 + 25}{2} = \frac{45}{2} \] Hence, \( P(x, y) = (5, \frac{45}{2}) \). Therefore, Rashmi should post her blue flag at \( \frac{45}{2} = 22.5 \) m on the 5th line. |
Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6). | |
Let, the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6) is k:1, then by section formula: $$\left(\frac{kx_{2}+x_{1}}{k+1},\frac{ky_{2}+y_{1}}{k+1}\right)$$ Putting value $$(-1,6)=\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$$ $$-1=\left(\frac{6k-3}{k+1}\right)$$ $$-k-1=6k-3$$ $$-k-6k=-3+1$$ $$7k=2$$ $$k:1 = 2:7$$ $$6=\left(\frac{-8k+10}{k+1}\right)$$ $$6k+6=-8k+10$$ $$6k+8k=10-6$$ $$14k=4$$ both side divided by 2: $$7k=2$$ $$k:1 = 2:7$$ |
Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. | |
Let the ratio in which the line segment joining points A(1, –5) and B(–4, 5) is divided by the x-axis be k:1. Therefore, the coordinates of the point of division can be denoted as P(x, y) and can be written as: $$\left(\frac{-4k+1}{k+1},\frac{5k-5}{k+1}\right)$$ Since any point on the x-axis has a y-coordinate of 0, we set the y-coordinate expression to 0: $$\frac{5k-5}{k+1}=0$$ $$5k=5$$ $$k=1$$ Thus, the x-axis divides the line segment in the ratio 1:1. Now, we find the coordinates of the point of division, P(x, y): $$P(x,y)=\left(\frac{-4(1)+1}{1+1},\frac{5(1)-5}{1+1}\right)$$ $$P(x,y)=\left(\frac{-4+1}{2},\frac{5-5}{2}\right)$$ $$P(x,y)=\left(\frac{-3}{2},\frac{0}{2}\right)$$ $$P(x,y)=\left(\frac{-3}{2},0\right)$$ Therefore, the x-axis divides the line segment joining A and B in the ratio 1:1, and the coordinates of the point of division, P, are \(\left(\frac{-3}{2},0\right)\) |
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. | |
We know that the midpoint formula for a line segment with endpoints \(\left(x_{1},y_{1}\right)\) and \(\left(x_{2},y_{2}\right)\) is: $$Mid Point=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$$ Given points : A(1, 2), B(4, y), C(x, 6) and D(3, 5), So Midpoint of AC: $$\left(\frac{1+x}{2},\frac{2+6}{2}\right)=\left(\frac{1+x}{2},4\right)$$ Midpoint of BD: $$\left(\frac{4+3}{2},\frac{y+5}{2}\right)=\left(\frac{7}{2},\frac{y+5}{2}\right)$$ We also know that in a parallelogram, the midpoint of one diagonal is the same as the midpoint of the other diagonal. So, $$\left(\frac{1+x}{2},4\right)=\left(\frac{7}{2},\frac{y+5}{2}\right)$$ Now, equate corresponding coordinates: For \(x\): $$\frac{1+x}{2}=\frac{7}{2}$$ $$2+2x=14$$ $$2x=14-2$$ $$2x=12$$ $$x=6$$ For \(y\): $$4=\frac{y+5}{2}$$ $$8=y+5$$ $$8-5=y$$ $$y=3$$ So, the values of \(x\) and \(y\) that satisfy the conditions are \(x=6\) and \(y=3\). |
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4). | |
Given: Midpoint of AB is (2,—3) Coordinate of B is (1,4) Let the coordinate of point A is (x,y) We know that midpoint is: $$Mid Point=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$$ Putting value $$(2,-3)=\left(\frac{x+1}{2},\frac{y+4}{2}\right)$$ Now, equate corresponding coordinates: For \(x\): $$2=\frac{x+1}{2}$$ $$4=x+1$$ $$4-1=x$$ $$x=3$$ For \(y\): $$-3=\frac{y+4}{2}$$ $$-6=y+4$$ $$-6-4=y$$ $$y=-10$$ Hence, The coordinates of \(A(3,-10)\). |
If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that \(AP=\frac{3}{7}AB\) and P lies on the line segment AB | |
Given that \(A(-2, -2)\) and \(B(2, -4)\). The vector \(\vec{AB}\) is given by: \[ \vec{AB} = (2 - (-2), (-4) - (-2)) = (4, -2) \] As per given \(AP = \frac{3}{7} AB\), so: \[ \vec{AP} = \frac{3}{7} \times \vec{AB} \] \[ \vec{AP} = \frac{3}{7} \times 4,\frac{3}{7} \times (-2) = \left(\frac{12}{7}, -\frac{6}{7}\right) \] The coordinates of point P, \( (x, y) \), are given by: \[ (x, y) = (x + \frac{12}{7}, y - \frac{6}{7}) \] \[ (x, y) = (-2 + \frac{12}{7}, -2 - \frac{6}{7}) \] \[ (x, y) = \left(\frac{-2 \times 7 + 12}{7}, \frac{-2 \times 7 - 6}{7}\right) \] \[ (x, y) = \left(\frac{-14 + 12}{7}, \frac{-20}{7}\right) \] \[ (x, y) = \left(\frac{-2}{7}, -\frac{20}{7}\right) \] Therefore, the coordinates of point P are \((\frac{-2}{7}, -\frac{20}{7})\). Alternative method: Given : The coordinates of point A and B are (-2,-2) and (2,-4). Since \(AP=\frac{3}{7}AB\) Therefore, Ratio between AP and PB = 3:4 So, Point P divides the line segment AB in the ratio 3:4. Then Coordinate of P is: $$\left( x = \frac{m_1 \cdot x_2 + m_2 \cdot x_1}{m_1 + m_2}, y = \frac{m_1 \cdot y_2 + m_2 \cdot y_1}{m_1 + m_2}\right)$$ putting value $$\left( x = \frac{3(2) + 4(-2)}{3 +4}, y = \frac{3(-4) + 4(-2)}{3 + 4}\right)$$ $$\left( x = \frac{6-8}{7}, y = \frac{-12-8}{7}\right)$$ $$\left( x = -\frac{7}{2}, y = -\frac{20}{7}\right)$$ Therefore, the coordinates of point P are \((-\frac{2}{7}, -\frac{20}{7})\). |
Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts. | ||
Given: the coordinates of the points that divide the line segment joining \(A(-2, 2)\) and \(B(2, 8)\) into four equal parts. Let the points dividing the line segment be \(P_1\), \(P_2\), and \(P_3\). The section formula for dividing a line segment into \(n\) equal parts is: \[ P_i = \left(\frac{{(n - i)x_1 + ix_2}}{{n}}, \frac{{(n - i)y_1 + iy_2}}{{n}}\right) \] where \(i = 1, 2, \ldots, n-1\). For dividing the segment AB into four equal parts, \(n = 4\), and we have three points to find (\(P_1\), \(P_2\), and \(P_3\)). 1. Coordinates of \(P_1\): \[ P_1 = \left(\frac{{3(-2) + 1(2)}}{4}, \frac{{3(2) + 1(8)}}{4}\right) = (-1, 3.5) \] 2. Coordinates of \(P_2\): \[ P_2 = \left(\frac{{2(-2) + 2(2)}}{4}, \frac{{2(2) + 2(8)}}{4}\right) = (0, 5) \] 3. Coordinates of \(P_3\): \[ P_3 = \left(\frac{{1(-2) + 3(2)}}{4}, \frac{{1(2) + 3(8)}}{4}\right) = (1, 6.5) \] Therefore, the coordinates of the points dividing the line segment AB into four equal parts are \(P_1(-1, 3.5)\), \(P_2(0, 5)\), and \(P_3(1, 6.5)\). Alternative Method:
As per figure, we can be observed that points P, Q, R are dividing the line segment AB in a ratio 1:3, 1:1, 3:1, respectively. Coordinate of P: \[P=\left(\frac{1(2)+3(-2)}{1+3},\frac{1(8)+3(2)}{1+3}\right)\] \[P=\left(\frac{2-6}{4},\frac{8+6}{4}\right)\] \[P=\left(\frac{-4}{4},\frac{14}{4}\right)\] \[P=\left(-1,\frac{7}{2}\right)=\left(-1,3.5\right)\] Coordinate of Q: \[Q=\left(\frac{2(1)+2(1)}{1+1},\frac{2(1)+8(1)}{1+1}\right)\] \[Q=\left(\frac{2-2}{2},\frac{2+8}{2}\right)\] \[Q=\left(\frac{0}{2},\frac{10}{2}\right)\] \[Q=\left(0,5\right)\] Coordinate of R: \[R=\left(\frac{3(2)+1(-2)}{3+1},\frac{3(8)+1(2)}{3+1}\right)\] \[R=\left(\frac{6-2}{4},\frac{24+2}{4}\right)\] \[R=\left(\frac{4}{4},\frac{26}{4}\right)\] \[R=\left(0,\frac{13}{2}\right)=\left(0,6.5\right)\] Therefore, the coordinates of the points dividing the line segment AB into four equal parts are \(P_1(-1, 3.5)\), \(P_2(0, 5)\), and \(P_3(1, 6.5)\). |
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint : Area of a rhombus\(=\frac{1}{2}\) (product of its diagonals)]. | |
| The vertices of the rhombus are given as \(A(3, 0)\), \(B(4, 5)\), \(C(-1, 4)\), and \(D(-2, -1)\). The formula for the area of a rhombus using the lengths of its diagonals is: \[ \text{Area} = \frac{1}{2} \times \text{product of diagonals} \] By the distance formula: Length of diagonal \(AC\): \[AC=\sqrt{\left(x_{C}-x_{A}\right)+\left(y_{C}-y_{A}\right)}\] \[ AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} \] \[AC=\sqrt{(-4)^{2}+4^{2}}=\sqrt{16+16}=\sqrt{32}\] Length of diagonal \(BD\): \[BD=\sqrt{\left(x_{D}-x_{B}\right)+\left(y_{D}-y_{B}\right)}\] \[ BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{72} \] \[BD=\sqrt{(-6)^{2}+(-6)^{2}}=\sqrt{36+36}=\sqrt{72}\] Now, calculate the product of the diagonals: \[ \text{Area} = \frac{1}{2} \times \sqrt{32} \times \sqrt{72} \] \[ \text{Area} = \frac{1}{2} \times \sqrt{32 \times 72} \] \[ \text{Area} = \frac{1}{2} \times \sqrt{2304} \] \[ \text{Area} = \frac{1}{2} \times 48=24 \] Therefore, the area of the rhombus is \(24\) square units. |
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