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Samacheer Kalvi 10th Maths Unit Exercise 5 Solutions

10th Maths Unit Exercise 5 solutions (Chapter 5 Coordinate Geometry) with step-by-step answers for all 10 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance9 answers · full steps below
  1. Q1ABCDABCD is a rhombus (all sides =612=\dfrac{\sqrt{61}}{2}, but diagonals 6≠56\neq 5).
  2. Q2The third vertex is (72,132)\left(\dfrac72,\dfrac{13}{2}\right) or (−32,32)\left(-\dfrac32,\dfrac32\right).
  3. Q3The three lines are concurrent at (1,−1)(1,-1), so the area of the triangle formed is 00 sq.units.
  4. Q4k=−5k=-5
  5. Q62x−3y−6=02x-3y-6=0 and 3x−2y+6=03x-2y+6=0
  6. Q7He could sell 13401340 litres per week at ₹1717/litre.
  7. Q8The image is (−1,−4)(-1,-4).
  8. Q913x+13y−6=013x+13y-6=0
  9. Q10119x+102y−125=0119x+102y-125=0

Q1. PQRS is a rectangle formed by joining the points P(−1,−1)P(-1,-1) , Q(−1,4)Q(-1,4) …

PQRS is a rectangle formed by joining the points P(−1,−1)P(-1,-1), Q(−1,4)Q(-1,4), R(5,4)R(5,4) and S(5,−1)S(5,-1). AA, BB, CC and DD are the mid-points of PQPQ, QRQR, RSRS and SPSP respectively. Is the quadrilateral ABCDABCD a square, a rectangle or a rhombus? Justify your answer.

Show steps (11)

  1. Mid-points: A=(−1−12,−1+42)=(−1,32)A=\left(\dfrac{-1-1}{2},\dfrac{-1+4}{2}\right)=\left(-1,\dfrac{3}{2}\right)

  2. B=(−1+52,4+42)=(2,4)B=\left(\dfrac{-1+5}{2},\dfrac{4+4}{2}\right)=(2,4)

  3. C=(5+52,4−12)=(5,32)C=\left(\dfrac{5+5}{2},\dfrac{4-1}{2}\right)=\left(5,\dfrac{3}{2}\right)

  4. D=(5−12,−1−12)=(2,−1)D=\left(\dfrac{5-1}{2},\dfrac{-1-1}{2}\right)=(2,-1)

  5. AB=(2+1)2+(4−32)2=9+254=612AB=\sqrt{(2+1)^2+\left(4-\dfrac32\right)^2}=\sqrt{9+\dfrac{25}{4}}=\dfrac{\sqrt{61}}{2}

  6. BC=(5−2)2+(32−4)2=612BC=\sqrt{(5-2)^2+\left(\dfrac32-4\right)^2}=\dfrac{\sqrt{61}}{2}

  7. CD=(2−5)2+(−1−32)2=612CD=\sqrt{(2-5)^2+\left(-1-\dfrac32\right)^2}=\dfrac{\sqrt{61}}{2}

  8. DA=(−1−2)2+(32+1)2=612DA=\sqrt{(-1-2)^2+\left(\dfrac32+1\right)^2}=\dfrac{\sqrt{61}}{2}

  9. All four sides are equal.

  10. Diagonals: AC=(5+1)2+0=6AC=\sqrt{(5+1)^2+0}=6, BD=0+(4+1)2=5BD=\sqrt{0+(4+1)^2}=5

  11. Since the diagonals are not equal, ABCDABCD is not a square or a rectangle.

Answer

ABCDABCD is a rhombus (all sides =612=\dfrac{\sqrt{61}}{2}, but diagonals 6≠56\neq 5).

Q2. The area of a triangle is 55 sq.units. Two of its vertices …

The area of a triangle is 55 sq.units. Two of its vertices are (2,1)(2,1) and (3, −2)(3,\,-2). The third vertex is (x, y)(x,\,y) where y=x+3y = x + 3. Find the coordinates of the third vertex.

Show steps (8)

  1. Vertices: (2,1)(2,1), (3,−2)(3,-2), (x, x+3)(x,\,x+3).

  2. Area =12{(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)}=\dfrac12\{(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\}

  3. x1y2+x2y3+x3y1=2(−2)+3(x+3)+x(1)=4x+5x_1y_2+x_2y_3+x_3y_1=2(-2)+3(x+3)+x(1)=4x+5

  4. x2y1+x3y2+x1y3=3(1)+x(−2)+2(x+3)=9x_2y_1+x_3y_2+x_1y_3=3(1)+x(-2)+2(x+3)=9

  5. Area =12∣4x+5−9∣=12∣4x−4∣=5=\dfrac12|4x+5-9|=\dfrac12|4x-4|=5

  6. ∣4x−4∣=10⇒4x−4=10|4x-4|=10\Rightarrow 4x-4=10 or 4x−4=−104x-4=-10

  7. x=72x=\dfrac72 or x=−32x=-\dfrac32

  8. y=x+3y=x+3 gives y=132y=\dfrac{13}{2} or y=32y=\dfrac32

Answer

The third vertex is (72,132)\left(\dfrac72,\dfrac{13}{2}\right) or (−32,32)\left(-\dfrac32,\dfrac32\right).

Q3. Find the area of a triangle formed by the lines 3x+y−2=03x + y - 2 = 0 …

Find the area of a triangle formed by the lines 3x+y−2=03x + y - 2 = 0, 5x+2y−3=05x + 2y - 3 = 0 and 2x−y−3=02x - y - 3 = 0

Show steps (6)

  1. Solve 3x+y−2=03x+y-2=0 and 5x+2y−3=05x+2y-3=0.

  2. From the first, y=2−3xy=2-3x. Then 5x+2(2−3x)−3=−x+1=0⇒x=1, y=−15x+2(2-3x)-3=-x+1=0\Rightarrow x=1,\ y=-1.

  3. So the first two lines meet at (1,−1)(1,-1).

  4. Check (1,−1)(1,-1) in the third line: 2(1)−(−1)−3=2+1−3=02(1)-(-1)-3=2+1-3=0.

  5. So (1,−1)(1,-1) lies on all three lines — the lines are concurrent.

  6. Three concurrent lines do not enclose a triangle.

Answer

The three lines are concurrent at (1,−1)(1,-1), so the area of the triangle formed is 00 sq.units.

Q4. If vertices of a quadrilateral are at A(−5,7)A(-5,7) , B(−4,k)B(-4,k) , C(−1,−6)C(-1,-6) …

If vertices of a quadrilateral are at A(−5,7)A(-5,7), B(−4,k)B(-4,k), C(−1,−6)C(-1,-6) and D(4,5)D(4,5) and its area is 7272 sq.units. Find the value of kk.

Show steps (8)

  1. Area =12{(x1y2+x2y3+x3y4+x4y1)−(x2y1+x3y2+x4y3+x1y4)}=\dfrac12\{(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4)\}

  2. With A(−5,7),B(−4,k),C(−1,−6),D(4,5)A(-5,7),B(-4,k),C(-1,-6),D(4,5):

  3. First bracket =(−5)k+(−4)(−6)+(−1)(5)+4(7)=−5k+47=(-5)k+(-4)(-6)+(-1)(5)+4(7)=-5k+47

  4. Second bracket =(−4)(7)+(−1)k+4(−6)+(−5)(5)=−k−77=(-4)(7)+(-1)k+4(-6)+(-5)(5)=-k-77

  5. Area =12∣(−5k+47)−(−k−77)∣=12∣−4k+124∣=72=\dfrac12|(-5k+47)-(-k-77)|=\dfrac12|{-4k+124}|=72

  6. ∣−4k+124∣=144|-4k+124|=144

  7. −4k+124=144⇒k=−5-4k+124=144\Rightarrow k=-5

  8. −4k+124=−144⇒k=67-4k+124=-144\Rightarrow k=67, but then the vertices A,B,C,DA,B,C,D are not in order (the quadrilateral crosses itself), so it is rejected.

Answer

k=−5k=-5

Q5. Without using distance formula, show that the points (−2,−1)(-2,-1) , (4,0)(4,0) …

Without using distance formula, show that the points (−2,−1)(-2,-1), (4,0)(4,0), (3,3)(3,3) and (−3,2)(-3,2) are vertices of a parallelogram.

Show steps (8)

  1. Let A(−2,−1)A(-2,-1), B(4,0)B(4,0), C(3,3)C(3,3), D(−3,2)D(-3,2).

  2. Slope of AB=0−(−1)4−(−2)=16AB=\dfrac{0-(-1)}{4-(-2)}=\dfrac16

  3. Slope of DC=3−23−(−3)=16DC=\dfrac{3-2}{3-(-3)}=\dfrac16

  4. Slope of AB=AB= slope of DCDC, so AB∥DCAB\parallel DC.

  5. Slope of BC=3−03−4=−3BC=\dfrac{3-0}{3-4}=-3

  6. Slope of AD=2−(−1)−3−(−2)=3−1=−3AD=\dfrac{2-(-1)}{-3-(-2)}=\dfrac{3}{-1}=-3

  7. Slope of BC=BC= slope of ADAD, so BC∥ADBC\parallel AD.

  8. Both pairs of opposite sides are parallel.

Answer

Since AB∥DCAB\parallel DC and BC∥ADBC\parallel AD, the points (−2,−1),(4,0),(3,3),(−3,2)(-2,-1),(4,0),(3,3),(-3,2) form a parallelogram.

Q6. Find the equations of the lines, whose sum and product of intercepts …

Find the equations of the lines, whose sum and product of intercepts are 11 and −6-6 respectively.

  1. Let the intercepts be aa and bb, with a+b=1a+b=1 and ab=−6ab=-6.

  2. a,ba,b are roots of t2−t−6=0t^2-t-6=0

  3. (t−3)(t+2)=0⇒t=3, −2(t-3)(t+2)=0\Rightarrow t=3,\,-2

  4. Case 1: a=3, b=−2a=3,\ b=-2. Then x3+y−2=1⇒2x−3y=6\dfrac{x}{3}+\dfrac{y}{-2}=1\Rightarrow 2x-3y=6

  5. Case 2: a=−2, b=3a=-2,\ b=3. Then x−2+y3=1⇒3x−2y+6=0\dfrac{x}{-2}+\dfrac{y}{3}=1\Rightarrow 3x-2y+6=0

Answer

2x−3y−6=02x-3y-6=0 and 3x−2y+6=03x-2y+6=0

Q7. The owner of a milk store finds that, he can sell 980980 …

The owner of a milk store finds that, he can sell 980980 litres of milk each week at ₹1414/litre and 12201220 litres of milk each week at ₹1616/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹1717/litre?

  1. Let xx = price per litre, yy = litres sold. Given points (14,980)(14,980) and (16,1220)(16,1220).

  2. Slope =1220−98016−14=2402=120=\dfrac{1220-980}{16-14}=\dfrac{240}{2}=120

  3. Equation: y−980=120(x−14)y-980=120(x-14)

  4. y=120x−1680+980=120x−700y=120x-1680+980=120x-700

  5. Put x=17x=17: y=120(17)−700=2040−700=1340y=120(17)-700=2040-700=1340

Answer

He could sell 13401340 litres per week at ₹1717/litre.

Q8. Find the image of the point (3,8)(3,8) with respect to the line …

Find the image of the point (3,8)(3,8) with respect to the line x+3y=7x + 3y = 7 assuming the line to be a plane mirror.

Show steps (7)

  1. Let the image be (a,b)(a,b) and the mirror line be x+3y=7x+3y=7.

  2. Slope of the mirror =−13=-\dfrac13, so slope of the line joining (3,8)(3,8) and (a,b)(a,b) is 33.

  3. b−8a−3=3⇒b=3a−1\dfrac{b-8}{a-3}=3\Rightarrow b=3a-1 ... (1)

  4. The mid-point (3+a2,8+b2)\left(\dfrac{3+a}{2},\dfrac{8+b}{2}\right) lies on the mirror:

  5. 3+a2+3(8+b2)=7⇒a+3b=−13\dfrac{3+a}{2}+3\left(\dfrac{8+b}{2}\right)=7\Rightarrow a+3b=-13 ... (2)

  6. Sub (1) in (2): a+3(3a−1)=−13⇒10a=−10⇒a=−1a+3(3a-1)=-13\Rightarrow 10a=-10\Rightarrow a=-1

  7. b=3(−1)−1=−4b=3(-1)-1=-4

Answer

The image is (−1,−4)(-1,-4).

Q9. Find the equation of a line passing through the point of intersection …

Find the equation of a line passing through the point of intersection of the lines 4x+7y−3=04x + 7y - 3 = 0 and 2x−3y+1=02x - 3y + 1 = 0 that has equal intercepts on the axes.

Show steps (8)

  1. Solve 4x+7y−3=04x+7y-3=0 and 2x−3y+1=02x-3y+1=0.

  2. Multiply the second by 22: 4x−6y+2=04x-6y+2=0

  3. Subtracting: 13y−5=0⇒y=51313y-5=0\Rightarrow y=\dfrac{5}{13}

  4. 2x=3(513)−1=213⇒x=1132x=3\left(\dfrac{5}{13}\right)-1=\dfrac{2}{13}\Rightarrow x=\dfrac{1}{13}

  5. Point of intersection: (113,513)\left(\dfrac{1}{13},\dfrac{5}{13}\right)

  6. Equal intercepts: xa+ya=1⇒x+y=a\dfrac{x}{a}+\dfrac{y}{a}=1\Rightarrow x+y=a

  7. a=113+513=613a=\dfrac{1}{13}+\dfrac{5}{13}=\dfrac{6}{13}

  8. So x+y=613x+y=\dfrac{6}{13}

Answer

13x+13y−6=013x+13y-6=0

Q10. A person standing at a junction (crossing) of two straight paths represented …

A person standing at a junction (crossing) of two straight paths represented by the equations 2x−3y+4=02x - 3y + 4 = 0 and 3x+4y−5=03x + 4y - 5 = 0 seek to reach the path whose equation is 6x−7y+8=06x - 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

Show steps (10)

  1. Least time means the shortest path, i.e. the perpendicular from the junction to the line 6x−7y+8=06x-7y+8=0.

  2. Junction: solve 2x−3y+4=02x-3y+4=0 and 3x+4y−5=03x+4y-5=0.

  3. 4×4\timesfirst: 8x−12y+16=08x-12y+16=0; 3×3\timessecond: 9x+12y−15=09x+12y-15=0

  4. Adding: 17x+1=0⇒x=−11717x+1=0\Rightarrow x=-\dfrac{1}{17}

  5. 2(−117)−3y+4=0⇒3y=6617⇒y=22172\left(-\dfrac{1}{17}\right)-3y+4=0\Rightarrow 3y=\dfrac{66}{17}\Rightarrow y=\dfrac{22}{17}

  6. Junction =(−117,2217)=\left(-\dfrac{1}{17},\dfrac{22}{17}\right)

  7. Slope of 6x−7y+8=06x-7y+8=0 is 67\dfrac{6}{7}, so the required slope =−76=-\dfrac76

  8. y−2217=−76(x+117)y-\dfrac{22}{17}=-\dfrac76\left(x+\dfrac{1}{17}\right)

  9. 6(17y−22)=−7(17x+1)6(17y-22)=-7(17x+1)

  10. 102y−132=−119x−7102y-132=-119x-7

Answer

119x+102y−125=0119x+102y-125=0

Quick check

  1. The point of intersection of 3x−y=43x - y = 4 and x+y=8x + y = 8 is

    1. (1) (5,3)(5,3)
    2. (2) (2,4)(2,4)
    3. (3) (3,5)(3,5)
    4. (4) (4,4)(4,4)
    Show answer

    Answer: (3) (3,5)(3,5)

    1. 3x−y=43x-y=4 and x+y=8x+y=8.

    2. Adding: 4x=12⇒x=34x=12\Rightarrow x=3.

    3. Then 3+y=8⇒y=53+y=8\Rightarrow y=5.

    4. Point of intersection is (3,5)(3,5).

    Question 1 of 5

  2. When proving that a quadrilateral is a trapezium, it is necessary to show

    1. (1) Two sides are parallel.
    2. (2) Two parallel and two non-parallel sides.
    3. (3) Opposite sides are parallel.
    4. (4) All sides are of equal length.
    Show answer

    Answer: (2) Two parallel and two non-parallel sides.

    1. A trapezium has exactly one pair of parallel sides.

    2. So we must show two sides are parallel and the other two are not parallel.

    Question 2 of 5

  3. If slope of the line PQPQ is 13\dfrac{1}{\sqrt{3}} then slope of the perpendicular bisector of PQPQ is

    1. (1) 3\sqrt{3}
    2. (2) −3-\sqrt{3}
    3. (3) 13\dfrac{1}{\sqrt{3}}
    4. (4) 00
    Show answer

    Answer: (2) −3-\sqrt{3}

    1. The perpendicular bisector of PQPQ is perpendicular to PQPQ.

    2. Slope of PQ=13PQ=\dfrac{1}{\sqrt{3}}.

    3. Required slope =−1(1/3)=−3=-\dfrac{1}{(1/\sqrt{3})}=-\sqrt{3}.

    Question 3 of 5

  4. The straight line given by the equation x=11x = 11 is

    1. (1) parallel to XX axis
    2. (2) parallel to YY axis
    3. (3) passing through the origin
    4. (4) passing through the point (0,11)(0,11)
    Show answer

    Answer: (2) parallel to YY axis

    1. x=11x=11 means every point on it has xx coordinate 1111, whatever be yy.

    2. Such a line is vertical, hence parallel to the Y axis.

    Question 4 of 5

  5. Consider four straight lines (i) l1; 3y=4x+5l_1;\ 3y = 4x + 5 (ii) l2; 4y=3x−1l_2;\ 4y = 3x - 1 (iii) l3; 4y+3x=7l_3;\ 4y + 3x = 7 (iv) l4; 4x+3y=2l_4;\ 4x + 3y = 2 Which of the following statement is true ?

    1. (1) l1l_1 and l2l_2 are perpendicular
    2. (2) l1l_1 and l4l_4 are parallel
    3. (3) l2l_2 and l4l_4 are perpendicular
    4. (4) l2l_2 and l3l_3 are parallel
    Show answer

    Answer: (3) l2l_2 and l4l_4 are perpendicular

    1. Slopes: l1=43l_1=\dfrac{4}{3}, l2=34l_2=\dfrac{3}{4}, l3=−34l_3=-\dfrac{3}{4}, l4=−43l_4=-\dfrac{4}{3}.

    2. m2×m4=34×(−43)=−1m_2\times m_4=\dfrac{3}{4}\times\left(-\dfrac{4}{3}\right)=-1.

    3. So l2l_2 and l4l_4 are perpendicular.

    Question 5 of 5
Chapter 5 formulas
  • Area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃)

    Area=12{(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)} sq.units\text{Area}=\frac{1}{2}\left\{(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\right\}\ \text{sq.units}

  • Collinearity condition (area)

    Area of △ABC=0orx1y2+x2y3+x3y1=x2y1+x3y2+x1y3\text{Area of }\triangle ABC=0\quad\text{or}\quad x_1y_2+x_2y_3+x_3y_1=x_2y_1+x_3y_2+x_1y_3

  • Collinearity condition (slope)

    slope of AB=slope of BC=slope of AC\text{slope of }AB=\text{slope of }BC=\text{slope of }AC

  • Area of a quadrilateral with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃), (x₄,y₄)

    Area=12{(x1y2+x2y3+x3y4+x4y1)−(x2y1+x3y2+x4y3+x1y4)} sq.units\text{Area}=\frac{1}{2}\left\{(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4)\right\}\ \text{sq.units}

  • Slope of a line making angle θ with the positive direction of the X axis

    m=tan⁡θm=\tan\theta

  • Slope of a line joining two distinct points A(x₁,y₁) and B(x₂,y₂)

    m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}

  • Slope of the line ax + by + c = 0

    m=−abm=\frac{-a}{b}

  • General form of a straight line

    ax+by+c=0ax+by+c=0

  • Point-slope form

    y−y1=m(x−x1)y-y_1=m(x-x_1)

  • Slope-intercept form

    y=mx+cy=mx+c

  • Two point form

    y−y1y2−y1=x−x1x2−x1\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}

  • Intercept form

    xa+yb=1\frac{x}{a}+\frac{y}{b}=1

  • Line parallel to the Y axis

    x=cx=c

  • Line parallel to the X axis

    y=by=b

  • Condition for two straight lines to be parallel

    m1=m2m_1=m_2

  • Condition for two straight lines to be perpendicular

    m1×m2=−1m_1\times m_2=-1

Answers checked against the Samacheer Kalvi textbook · Updated

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