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EAPCET Mathematics · Intermediate first year · Paper IB

The Straight Line

The Straight Line covers every form of a line’s equation, the angle between lines, distances, concurrency and families of lines.

Subject
Mathematics
Taught in
Intermediate first year · Paper IB
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • y−y1=m(x−x1)y - y_{1} = m(x - x_{1})
  • Angle between lines: tan⁡θ=∣m1−m21+m1m2∣\tan\theta = \left|\dfrac{m_{1} - m_{2}}{1 + m_{1}m_{2}}\right|
  • Distance of (x1,y1)(x_{1}, y_{1}) from ax+by+c=0ax + by + c = 0 is ∣ax1+by1+c∣a2+b2\dfrac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2} + b^{2}}}
  • Distance between parallel lines ax+by+c1=0ax + by + c_{1} = 0 and ax+by+c2=0ax + by + c_{2} = 0 is ∣c1−c2∣a2+b2\dfrac{|c_{1} - c_{2}|}{\sqrt{a^{2} + b^{2}}}
  • Three lines are concurrent when the determinant of their coefficients is 0
  • Foot of the perpendicular (h,k)(h, k) from (x1,y1)(x_{1}, y_{1}) to ax+by+c=0ax + by + c = 0: h−x1a=k−y1b=−ax1+by1+ca2+b2\dfrac{h - x_{1}}{a} = \dfrac{k - y_{1}}{b} = -\dfrac{ax_{1} + by_{1} + c}{a^{2} + b^{2}}
  • Image (h,k)(h, k) of (x1,y1)(x_{1}, y_{1}) in the line ax+by+c=0ax + by + c = 0: h−x1a=k−y1b=−2(ax1+by1+c)a2+b2\dfrac{h - x_{1}}{a} = \dfrac{k - y_{1}}{b} = -\dfrac{2(ax_{1} + by_{1} + c)}{a^{2} + b^{2}}
  • Bisectors of the angles between a1x+b1y+c1=0a_{1}x + b_{1}y + c_{1} = 0 and a2x+b2y+c2=0a_{2}x + b_{2}y + c_{2} = 0: a1x+b1y+c1a12+b12=±a2x+b2y+c2a22+b22\dfrac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm\dfrac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}

What this chapter covers

  • Slope–intercept, point–slope, intercept and normal forms
  • Angle between two lines
  • Distance of a point from a line
  • Foot of the perpendicular and image of a point
  • Concurrent lines and families of lines

How it is usually asked

  • Write the equation of a line from given conditions
  • Find a distance, a foot of perpendicular or an image
  • Test concurrency or find the point of intersection
  • Find the area of a triangle formed by lines

Understanding The Straight Line

A straight line is fixed by its direction, measured by the slope m = tan θ, together with one point on it, or by equivalent information. That is why it has several forms: slope–intercept y = mx + c, point–slope y − y₁ = m(x − x₁), intercept form x/a + y/b = 1 and normal form x cos α + y sin α = p. All of them can be rearranged into the general form ax + by + c = 0, whose slope is −a/b.

Slopes decide the relative position of two lines: equal slopes mean parallel lines, a product of −1 means perpendicular lines, and otherwise tan θ = |(m₁ − m₂)/(1 + m₁m₂)| gives the acute angle between them. The perpendicular distance from (x₁, y₁) to ax + by + c = 0 is |ax₁ + by₁ + c|/√(a² + b²), and related formulas give the foot of the perpendicular and the image of a point.

The equation L₁ + λL₂ = 0 describes every line through the intersection of L₁ = 0 and L₂ = 0, which saves solving for the point. Three lines are concurrent when the determinant of their coefficients is zero. EAPCET questions ask you to build a line from given conditions, compute distances, feet and images, test concurrency or find areas of triangles formed by lines.

Worked example

EAPCET-style question

The distance between the parallel lines 3x+4y−7=03x + 4y - 7 = 0 and 6x+8y+1=06x + 8y + 1 = 0 is

  1. (1)32\dfrac{3}{2}
  2. (2)85\dfrac{8}{5}
  3. (3)65\dfrac{6}{5}
  4. (4)34\dfrac{3}{4}

Solution

  1. Make the xx and yy coefficients equal: divide the second line by 2 to get 3x+4y+12=03x + 4y + \tfrac{1}{2} = 0.
  2. Now c1=−7c_{1} = -7 and c2=12c_{2} = \tfrac{1}{2}, so the distance =∣c1−c2∣32+42= \dfrac{|c_{1} - c_{2}|}{\sqrt{3^{2} + 4^{2}}}.
  3. ∣−7−12∣=152\left|-7 - \tfrac{1}{2}\right| = \tfrac{15}{2}, and 25=5\sqrt{25} = 5.
  4. Distance =15/25=32= \dfrac{15/2}{5} = \dfrac{3}{2}.

Answer: 32\dfrac{3}{2}

Shortcuts and results to remember

  • The centroid GG lies on the line joining the circumcentre SS and the orthocentre HH and divides SHSH in the ratio 1:21 : 2.
  • The centroid divides every median in the ratio 2 : 1 measured from the vertex.
  • In a right-angled triangle the circumcentre is the midpoint of the hypotenuse and the orthocentre is the right-angled vertex.
  • In an equilateral triangle the centroid, orthocentre, circumcentre and incentre all coincide.

More practice questions

Practice question 1

The image of the point (3,4)(3, 4) in the line x+y−1=0x + y - 1 = 0 is

  1. (1)(0,1)(0, 1)
  2. (2)(−3,−2)(-3, -2)
  3. (3)(−2,−3)(-2, -3)
  4. (4)(3,−4)(3, -4)

Solution

  1. Use h−x1a=k−y1b=−2(ax1+by1+c)a2+b2\dfrac{h - x_{1}}{a} = \dfrac{k - y_{1}}{b} = -\dfrac{2(ax_{1} + by_{1} + c)}{a^{2} + b^{2}} with a=b=1a = b = 1, c=−1c = -1
  2. ax1+by1+c=3+4−1=6ax_{1} + by_{1} + c = 3 + 4 - 1 = 6, so the common ratio is −2(6)2=−6\dfrac{-2(6)}{2} = -6
  3. h=3−6=−3h = 3 - 6 = -3 and k=4−6=−2k = 4 - 6 = -2
  4. Check: the midpoint (0,1)(0, 1) lies on x+y−1=0x + y - 1 = 0

Answer: (−3,−2)(-3, -2)

Practice question 2

The circumcentre of the triangle with vertices (0,0)(0, 0), (4,0)(4, 0) and (0,6)(0, 6) is

  1. (1)(43,2)\left(\tfrac{4}{3}, 2\right)
  2. (2)(2,3)(2, 3)
  3. (3)(0,0)(0, 0)
  4. (4)(1,1)(1, 1)

Solution

  1. The sides along the axes are perpendicular, so the triangle is right-angled at the origin
  2. The circumcentre of a right-angled triangle is the midpoint of the hypotenuse
  3. Hypotenuse joins (4,0)(4, 0) and (0,6)(0, 6); its midpoint is (2,3)(2, 3)

Answer: (2,3)(2, 3)

Common mistakes to avoid

  • Using |c₁ − c₂|/√(a² + b²) before making the x and y coefficients of both lines identical, which gives a wrong value like 8/5.
  • Applying m₁m₂ = −1 when one line is vertical; a vertical and a horizontal line are perpendicular even though the vertical slope is undefined.
  • Reading intercepts from ax + by = c without dividing by c; the intercept form needs 1 on the right side.
  • Forgetting the modulus in the angle formula and reporting a negative tan θ; the formula with modulus gives the acute angle, and the obtuse one is π − θ.

How to prepare it

Choose the form that matches the data: intercepts suggest intercept form, a perpendicular from the origin suggests normal form.

The image of a point in a line is twice the foot of the perpendicular minus the point — learn it as one formula.

Once the formulas feel familiar, test them the way the exam does: practise The Straight Line questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise The Straight Line chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is The Straight Line part of the EAPCET Mathematics syllabus?
Yes. The Straight Line is taught in Intermediate first year · Paper IB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare The Straight Line for EAPCET?
Choose the form that matches the data: intercepts suggest intercept form, a perpendicular from the origin suggests normal form. The image of a point in a line is twice the foot of the perpendicular minus the point — learn it as one formula.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in The Straight Line?
Using |c₁ − c₂|/√(a² + b²) before making the x and y coefficients of both lines identical, which gives a wrong value like 8/5. Applying m₁m₂ = −1 when one line is vertical; a vertical and a horizontal line are perpendicular even though the vertical slope is undefined. Reading intercepts from ax + by = c without dividing by c; the intercept form needs 1 on the right side. Forgetting the modulus in the angle formula and reporting a negative tan θ; the formula with modulus gives the acute angle, and the obtuse one is π − θ.