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Samacheer Kalvi 10th Maths Exercise 5.5 Solutions

10th Maths Exercise 5.5 solutions (Chapter 5 Coordinate Geometry) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1(B) 2525 sq.units
  2. Q2(A) x=10x = 10
  3. Q3(B) parallel to YY axis
  4. Q4(C) 99
  5. Q5(C) (3,5)(3,5)
  6. Q6(D) 22
  7. Q7(B) 11
  8. Q8(B) −3-\sqrt{3}
  9. Q9(A) 8x+5y=408x + 5y = 40
  10. Q10(C) 3x+7y=03x + 7y = 0
  11. Q11(C) l2l_2 and l4l_4 are perpendicular
  12. Q12(A) The slope is 0.5 and the yy intercept is 2.6
  13. Q13(B) Two parallel and two non-parallel sides.
  14. Q14(B) The slopes of two pair of opposite sides
  15. Q15(B) x+y=3; 3x+y=7x + y = 3;\ 3x + y = 7

Q1. The area of triangle formed by the points (−5,0)(-5,0) , (0,−5)(0,-5) and …

The area of triangle formed by the points (−5,0)(-5,0), (0,−5)(0,-5) and (5,0)(5,0) is

  1. (A) 00 sq.units
  2. (B) 2525 sq.units
  3. (C) 55 sq.units
  4. (D) none of these

  1. The points (−5,0)(-5,0) and (5,0)(5,0) lie on the X axis, so the base =10=10 units.

  2. The third point (0,−5)(0,-5) is at a distance 55 from the X axis, so height =5=5 units.

  3. Area =12×10×5=25=\dfrac{1}{2}\times 10\times 5=25 sq.units.

Answer

(B) 2525 sq.units

Q2. A man walks near a wall, such that the distance between him …

A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the YY axis. The path travelled by the man is

  1. (A) x=10x = 10
  2. (B) y=10y = 10
  3. (C) x=0x = 0
  4. (D) y=0y = 0

  1. The wall is the Y axis, and his distance from it is always 1010 units.

  2. So his xx coordinate is always 1010.

  3. Path is the line x=10x=10.

Answer

(A) x=10x = 10

Q3. The straight line given by the equation x=11x = 11 is

  1. (A) parallel to XX axis
  2. (B) parallel to YY axis
  3. (C) passing through the origin
  4. (D) passing through the point (0,11)(0,11)

  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.

Answer

(B) parallel to YY axis

Q4. If (5,7)(5,7) , (3,p)(3,p) and (6,6)(6,6) are collinear, then the value of …

If (5,7)(5,7), (3,p)(3,p) and (6,6)(6,6) are collinear, then the value of pp is

  1. (A) 33
  2. (B) 66
  3. (C) 99
  4. (D) 1212

  1. Slope of the line joining (5,7)(5,7) and (6,6)(6,6) is 6−76−5=−1\dfrac{6-7}{6-5}=-1.

  2. For collinearity, slope of (5,7)(5,7) and (3,p)(3,p) is also −1-1.

  3. p−73−5=−1⇒p−7=2\dfrac{p-7}{3-5}=-1\Rightarrow p-7=2

  4. p=9p=9

Answer

(C) 99

Q5. The point of intersection of 3x−y=43x - y = 4 and x+y=8x + y = 8 is

  1. (A) (5,3)(5,3)
  2. (B) (2,4)(2,4)
  3. (C) (3,5)(3,5)
  4. (D) (4,4)(4,4)

  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).

Answer

(C) (3,5)(3,5)

Q6. The slope of the line joining (12,3)(12,3) , (4,a)(4,a) is 18\dfrac{1}{8} . …

The slope of the line joining (12,3)(12,3), (4,a)(4,a) is 18\dfrac{1}{8}. The value of 'aa' is

  1. (A) 11
  2. (B) 44
  3. (C) −5-5
  4. (D) 22

  1. Slope =a−34−12=a−3−8=\dfrac{a-3}{4-12}=\dfrac{a-3}{-8}.

  2. a−3−8=18⇒a−3=−1\dfrac{a-3}{-8}=\dfrac{1}{8}\Rightarrow a-3=-1

  3. a=2a=2

Answer

(D) 22

Q7. The slope of the line which is perpendicular to a line joining …

The slope of the line which is perpendicular to a line joining the points (0,0)(0,0) and (−8,8)(-8,8) is

  1. (A) −1-1
  2. (B) 11
  3. (C) 13\dfrac{1}{3}
  4. (D) −8-8

  1. Slope of the line joining (0,0)(0,0) and (−8,8)(-8,8) is 8−0−8−0=−1\dfrac{8-0}{-8-0}=-1.

  2. Product of slopes of perpendicular lines =−1=-1.

  3. Required slope =−1−1=1=\dfrac{-1}{-1}=1.

Answer

(B) 11

Q8. If slope of the line PQPQ is 13\dfrac{1}{\sqrt{3}} then slope of the …

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

  1. (A) 3\sqrt{3}
  2. (B) −3-\sqrt{3}
  3. (C) 13\dfrac{1}{\sqrt{3}}
  4. (D) 00

  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}.

Answer

(B) −3-\sqrt{3}

Q9. If AA is a point on the YY axis whose ordinate is …

If AA is a point on the YY axis whose ordinate is 8 and BB is a point on the XX axis whose abscissae is 5 then the equation of the line ABAB is

  1. (A) 8x+5y=408x + 5y = 40
  2. (B) 8x−5y=408x - 5y = 40
  3. (C) x=8x = 8
  4. (D) y=5y = 5

  1. AA is on the Y axis with ordinate 88, so A(0,8)A(0,8); BB is on the X axis with abscissa 55, so B(5,0)B(5,0).

  2. Intercept form: x5+y8=1\dfrac{x}{5}+\dfrac{y}{8}=1.

  3. Multiplying by 4040: 8x+5y=408x+5y=40.

Answer

(A) 8x+5y=408x + 5y = 40

Q10. The equation of a line passing through the origin and perpendicular to …

The equation of a line passing through the origin and perpendicular to the line 7x−3y+4=07x - 3y + 4 = 0 is

  1. (A) 7x−3y+4=07x - 3y + 4 = 0
  2. (B) 3x−7y+4=03x - 7y + 4 = 0
  3. (C) 3x+7y=03x + 7y = 0
  4. (D) 7x−3y=07x - 3y = 0

  1. Slope of 7x−3y+4=07x-3y+4=0 is −7−3=73\dfrac{-7}{-3}=\dfrac{7}{3}.

  2. Slope of the perpendicular line =−37=-\dfrac{3}{7}.

  3. Through the origin: y=−37x⇒3x+7y=0y=-\dfrac{3}{7}x\Rightarrow 3x+7y=0.

Answer

(C) 3x+7y=03x + 7y = 0

Q11. Consider four straight lines …

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. (A) l1l_1 and l2l_2 are perpendicular
  2. (B) l1l_1 and l4l_4 are parallel
  3. (C) l2l_2 and l4l_4 are perpendicular
  4. (D) l2l_2 and l3l_3 are parallel

  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.

Answer

(C) l2l_2 and l4l_4 are perpendicular

Q12. A straight line has equation 8y=4x+218y = 4x + 21 . Which of the following is …

A straight line has equation 8y=4x+218y = 4x + 21. Which of the following is true

  1. (A) The slope is 0.5 and the yy intercept is 2.6
  2. (B) The slope is 5 and the yy intercept is 1.6
  3. (C) The slope is 0.5 and the yy intercept is 1.6
  4. (D) The slope is 5 and the yy intercept is 2.6

  1. 8y=4x+21⇒y=48x+2188y=4x+21\Rightarrow y=\dfrac{4}{8}x+\dfrac{21}{8}.

  2. y=0.5x+2.625y=0.5x+2.625.

  3. Slope =0.5=0.5, yy intercept =2.625≈2.6=2.625\approx 2.6.

Answer

(A) The slope is 0.5 and the yy intercept is 2.6

Q13. When proving that a quadrilateral is a trapezium, it is necessary to …

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

  1. (A) Two sides are parallel.
  2. (B) Two parallel and two non-parallel sides.
  3. (C) Opposite sides are parallel.
  4. (D) All sides are of equal length.

  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.

Answer

(B) Two parallel and two non-parallel sides.

Q14. When proving that a quadrilateral is a parallelogram by using slopes you …

When proving that a quadrilateral is a parallelogram by using slopes you must find

  1. (A) The slopes of two sides
  2. (B) The slopes of two pair of opposite sides
  3. (C) The lengths of all sides
  4. (D) Both the lengths and slopes of two sides

  1. In a parallelogram both pairs of opposite sides are parallel.

  2. Using slopes, we find the slopes of the two pairs of opposite sides and show each pair is equal.

Answer

(B) The slopes of two pair of opposite sides

Q15. (2, 1)(2,\ 1) is the point of intersection of two lines.

  1. (A) x−y−3=0; 3x−y−7=0x - y - 3 = 0;\ 3x - y - 7 = 0
  2. (B) x+y=3; 3x+y=7x + y = 3;\ 3x + y = 7
  3. (C) 3x+y=3; x+y=73x + y = 3;\ x + y = 7
  4. (D) x+3y−3=0; x−y−7=0x + 3y - 3 = 0;\ x - y - 7 = 0

  1. Substitute x=2, y=1x=2,\ y=1 in option (B).

  2. x+y=2+1=3x+y=2+1=3 ✓

  3. 3x+y=6+1=73x+y=6+1=7 ✓

  4. Both are satisfied, so (2,1)(2,1) is their point of intersection.

Answer

(B) x+y=3; 3x+y=7x + y = 3;\ 3x + y = 7

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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