NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 7: Coordinate Geometry

Exercise 7.4

Step 1: Understand the problem.

We are given two points \((-4,3)\) and \((4,3)\). These are two corners (vertices) of an equilateral triangle. In an equilateral triangle, all sides are equal and all angles are \(60^\circ\).

We need to find the third vertex. There will be two possible positions for it, but only one of them will keep the origin \((0,0)\) inside the triangle.

Step 2: Find the length of the given side.

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by the distance formula:

\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]

Here, \((x_1,y_1) = (-4,3)\) and \((x_2,y_2) = (4,3)\).

So,

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

\[ = \sqrt{(8)^2 + 0^2} = \sqrt{64} = 8. \]

So, the side length of the equilateral triangle is \(s = 8\).

Step 3: Find the midpoint of the base.

The midpoint formula is:

\[ M = \Bigl(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Bigr). \]

Substituting values:

\[ M = \Bigl(\dfrac{-4 + 4}{2}, \dfrac{3 + 3}{2}\Bigr) = (0,3). \]

So the midpoint of the base is at \(M(0,3)\).

Step 4: Find the height of the equilateral triangle.

In an equilateral triangle, the altitude (height) can be found using the formula:

\[ h = \dfrac{\sqrt{3}}{2} \times s. \]

Substitute \(s = 8\):

\[ h = \dfrac{\sqrt{3}}{2} \times 8 = 4\sqrt{3}. \]

So the height is \(h = 4\sqrt{3}\).

Step 5: Find the possible third vertices.

The altitude passes through the midpoint \(M(0,3)\) and is perpendicular to the base. The base is horizontal (parallel to the x-axis), so the altitude is vertical (parallel to the y-axis).

That means the third vertex lies on the vertical line \(x = 0\), either above or below the base.

So the two possible points are:

\[ (0,3 + 4\sqrt{3}) \quad \text{(above)} \]

\[ (0,3 - 4\sqrt{3}) \quad \text{(below)} \]

Step 6: Choose the correct vertex using the condition.

We are told that the origin \((0,0)\) lies inside the triangle. If we choose the top point \((0,3 + 4\sqrt{3})\), the entire triangle will be above the x-axis, and the origin \((0,0)\) will lie outside.

But if we choose the bottom point \((0,3 - 4\sqrt{3})\), the triangle will stretch downward and include the origin \((0,0)\) inside it.

Final Answer:

The third vertex is \(\boxed{(0,\;3 - 4\sqrt{3})}\).

Step 1: Find the 4th vertex \(D\).

In a parallelogram, the rule is: \( \vec{A} + \vec{C} = \vec{B} + \vec{D} \).

So, \( D = A + C - B \).

Put the coordinates:

\( D = (6,1) + (9,4) - (8,2) \)

\( = (6+9-8,\ 1+4-2) = (7,3) \).

Step 2: Find midpoint \(E\) of \(DC\).

Midpoint formula: \( E = \Bigl(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\Bigr) \).

Here \(D(7,3), C(9,4)\).

\( E = \Bigl(\dfrac{7+9}{2},\dfrac{3+4}{2}\Bigr) = (8,3.5) \).

Step 3: Write vectors \(AD\) and \(AE\).

\( \overrightarrow{AD} = D - A = (7-6, 3-1) = (1,2) \).

\( \overrightarrow{AE} = E - A = (8-6, 3.5-1) = (2,2.5) \).

Step 4: Formula for area of triangle using vectors.

Area of triangle \(ADE = \dfrac{1}{2}\,\big| x_1y_2 - x_2y_1 \big| \).

Here \( (x_1,y_1)=(1,2),\ (x_2,y_2)=(2,2.5) \).

So, \( \text{Area} = \dfrac{1}{2}\,| (1)(2.5) - (2)(2) | \).

\( = \dfrac{1}{2}\,| 2.5 - 4 | \).

\( = \dfrac{1}{2}\,(1.5) = 0.75 = \dfrac{3}{4} \).

Final Answer: Area of triangle \(ADE = \dfrac{3}{4}\) square units.

(i) Finding point D

The median from A is a line that joins vertex A to the midpoint of side BC.

So, first we need the midpoint of BC. The midpoint formula is:

Midpoint = \(\Bigl(\dfrac{x_2+x_3}{2},\; \dfrac{y_2+y_3}{2}\Bigr)\)

Therefore, point D is:

\[ D\Bigl(\dfrac{x_2+x_3}{2},\; \dfrac{y_2+y_3}{2}\Bigr) \]

(ii) Finding point P on AD

P lies on the line segment AD such that the ratio \(AP:PD=2:1\).

That means P divides AD internally in the ratio 2:1.

For this, we use the section formula:

If a point divides a line joining points \((x_1,y_1)\) and \((x_2,y_2)\) in the ratio \(m:n\), then

\[ P = \Bigl(\dfrac{mx_2+nx_1}{m+n},\; \dfrac{my_2+ny_1}{m+n}\Bigr) \]

Here, point A is \((x_1,y_1)\), point D is \(\Bigl(\dfrac{x_2+x_3}{2},\;\dfrac{y_2+y_3}{2}\Bigr)\), and the ratio is 2:1.

So, applying the formula we get:

\[ P = \Bigl(\dfrac{2\cdot\dfrac{x_2+x_3}{2} + 1\cdot x_1}{3},\; \dfrac{2\cdot\dfrac{y_2+y_3}{2} + 1\cdot y_1}{3}\Bigr) \]

Simplifying:

\[ P = \Bigl(\dfrac{x_1+x_2+x_3}{3},\; \dfrac{y_1+y_2+y_3}{3}\Bigr) \]

(iii) Finding points Q and R

Now consider the other medians:

• Median from B goes to midpoint E of AC.
• Median from C goes to midpoint F of AB.

We repeat the same idea: divide each median in the ratio 2:1. This gives points Q and R.

When we apply the section formula, we will again get the same coordinates:

\[ Q = R = \Bigl(\dfrac{x_1+x_2+x_3}{3},\; \dfrac{y_1+y_2+y_3}{3}\Bigr) \]

(iv) Centroid of triangle

The centroid is the common point where all three medians meet.

So, the centroid \(G\) is also:

\[ G = \Bigl(\dfrac{x_1+x_2+x_3}{3},\; \dfrac{y_1+y_2+y_3}{3}\Bigr) \]

Final Note

This formula tells us: the centroid is simply the average of the x-coordinates and the average of the y-coordinates of the three vertices.

Step 1: Use the property of a parallelogram

In a parallelogram, the two diagonals bisect each other. This means the midpoints of both diagonals are the same.

Step 2: Write the coordinates of midpoints

Diagonal AC: Points are \(A(1,-2)\) and \(C(a,2)\).

Midpoint of AC:

\[ M_{AC} = \left(\dfrac{1+a}{2}, \dfrac{-2+2}{2}\right) = \left(\dfrac{1+a}{2}, 0\right). \]

Diagonal BD: Points are \(B(2,3)\) and \(D(-4,-3)\).

Midpoint of BD:

\[ M_{BD} = \left(\dfrac{2+(-4)}{2}, \dfrac{3+(-3)}{2}\right) = (-1, 0). \]

Step 3: Equating the midpoints

Since diagonals bisect each other, \(M_{AC} = M_{BD}\).

So, \(\dfrac{1+a}{2} = -1\).

Solve for \(a\):

\[ 1+a = -2 \;\Rightarrow\; a = -3. \]

Step 4: Find the base length AB

Points are \(A(1,-2)\) and \(B(2,3)\).

Distance formula:

\[ |AB| = \sqrt{(2-1)^2 + (3 - (-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}. \]

Step 5: Find the area of the parallelogram

Take vectors:

\(\overrightarrow{AB} = (2-1, 3-(-2)) = (1,5)\).

\(\overrightarrow{AD} = (-4-1, -3-(-2)) = (-5,-1)\).

Area of a parallelogram is the absolute value of the determinant of these two vectors:

\[ \text{Area} = |(1)(-1) - (5)(-5)| = |-1 + 25| = 24. \]

Step 6: Find the height when AB is the base

Area of parallelogram = base × height.

So, \( h = \dfrac{\text{Area}}{\text{Base}} = \dfrac{24}{\sqrt{26}}. \)

Final Answer:

\(a = -3\) and height on base \(AB = \dfrac{24}{\sqrt{26}}.\)

Step 1: To be equidistant from four points, Jaspal must stand at the circumcenter of the quadrilateral (if the four points lie on the same circle). The circumcenter is the point where the perpendicular bisectors of the sides meet.

Step 2: Take points \(A(3,5)\) and \(C(10,5)\). These two points are on the same horizontal line (since their \(y\)-coordinates are the same).

• The midpoint of \(AC\) is \(\bigl(\tfrac{3+10}{2},\tfrac{5+5}{2}\bigr)=(6.5,5)\).
• The line \(AC\) is horizontal, so its perpendicular bisector will be a vertical line.
• Equation of perpendicular bisector: \(x = 6.5\).

Step 3: Now consider \(B(6,8)\) and \(D(7,1)\).

• The midpoint of \(BD\) is \(\bigl(\tfrac{6+7}{2},\tfrac{8+1}{2}\bigr)=(6.5,4.5)\).
• Slope of \(BD\) = \(\tfrac{1-8}{7-6} = -7\).
• Slope of perpendicular bisector = negative reciprocal = \(-\tfrac{1}{-7}=\tfrac{1}{7}\).
• Equation of perpendicular bisector through midpoint: \(y-4.5=\tfrac{1}{7}(x-6.5)\).

Step 4: Find intersection of the two bisectors:

• From Step 2: \(x=6.5\).
• Substitute into equation of Step 3: \(y-4.5=\tfrac{1}{7}(6.5-6.5)=0\Rightarrow y=4.5\).
• So the intersection point is \((6.5,4.5)\).

Step 5: Check distances (all in SI units, each grid step is 1 unit):

• Distance to \(A(3,5)\): \(\sqrt{(6.5-3)^2+(4.5-5)^2}=\sqrt{(3.5)^2+(−0.5)^2}=\sqrt{12.25+0.25}=\sqrt{12.5}\).
• Distance to \(B(6,8)\): \(\sqrt{(6.5-6)^2+(4.5-8)^2}=\sqrt{0.25+12.25}=\sqrt{12.5}\).
• Distance to \(C(10,5)\): same calculation gives \(\sqrt{12.5}\).
• Distance to \(D(7,1)\): same calculation gives \(\sqrt{12.5}\).

Since all four distances are equal (about 3.54 units in SI), Jaspal can indeed be equidistant from all four points.

Final Answer: Jaspal should stand at position \(\boxed{(6.5,4.5)}\).

Step 1: Recall distance formula.

If two points are \((x_1,y_1)\) and \((x_2,y_2)\), then

\[ \,\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]

Step 2: Calculate distance from House to Bank (H to B).

H(2,4), B(5,8)

\[ HB = \sqrt{(5-2)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\;\text{km} \]

Step 3: Calculate distance from Bank to School (B to S).

B(5,8), S(13,14)

\[ BS = \sqrt{(13-5)^2 + (14-8)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\;\text{km} \]

Step 4: Calculate distance from School to Office (S to O).

S(13,14), O(13,26)

Here, the x-coordinates are the same, so the line is vertical.

\[ SO = |26 - 14| = 12\;\text{km} \]

Step 5: Add up the three parts.

\[ \text{Total distance via Bank and School} = HB + BS + SO = 5 + 10 + 12 = 27\;\text{km} \]

Step 6: Calculate direct distance from House to Office (H to O).

H(2,4), O(13,26)

\[ HO = \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{11^2 + 22^2} = \sqrt{121 + 484} = \sqrt{605}\;\text{km} \]

Using calculator: \(\sqrt{605} \approx 24.596\;\text{km}\).

Step 7: Find the extra distance travelled.

\[ \text{Extra distance} = (27 - \sqrt{605})\;\text{km} \approx (27 - 24.596)\;\text{km} = 2.404\;\text{km} \]

Final Answer: Ayush travels about 2.404 km more when he goes through Bank and School instead of going directly from House to Office.