SAT Weightage:

SAT — Coordinate Plane Strategy

SAT — Coordinate Plane Strategy — strategy and worked examples

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Chapter Overview & Weightage

Coordinate plane questions are guaranteed on every Digital SAT Math section. They test slopes, distance, midpoints, line equations, and basic conic sections (circles, parabolas).

Typical SAT weightage: 464-6 questions per Math section.

ConceptFrequency
Linear equations and slopesEvery test
Distance/midpoint formulasEvery test
Circle equationsMost tests
ParabolasSome tests
System of two linesEvery test

Key Concepts You Must Know

  • Slope formula: m=(y2y1)/(x2x1)m = (y_2 - y_1)/(x_2 - x_1)
  • Slope-intercept form y=mx+by = mx + b vs point-slope form
  • Parallel lines have equal slopes; perpendicular lines have slopes whose product is 1-1
  • Distance formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
  • Midpoint formula: M=((x1+x2)/2,(y1+y2)/2)M = ((x_1+x_2)/2, (y_1+y_2)/2)
  • Circle equation: (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2
  • Parabola in vertex form: y=a(xh)2+ky = a(x-h)^2 + k

Important Formulas

Slope-intercept: y=mx+by = mx + b

Point-slope: yy1=m(xx1)y - y_1 = m(x - x_1)

Standard: Ax+By=CAx + By = C, slope =A/B= -A/B

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

M=(x1+x22,y1+y22)M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)

(xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2

centre (h,k)(h, k), radius rr. Complete the square if given general form.

Solved Previous Year Questions

PYQ 1 (Practice Test)

The line y=3x5y = 3x - 5 is perpendicular to which line?

A line perpendicular to slope 33 has slope 1/3-1/3. Look for any line with slope 1/3-1/3, e.g., y=x/3+2y = -x/3 + 2.

PYQ 2 (Practice Test)

A circle has equation x2+y26x+4y12=0x^2 + y^2 - 6x + 4y - 12 = 0. Find the centre and radius.

Complete the square: (x26x+9)+(y2+4y+4)=12+9+4=25(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 = 25.

So (x3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25. Centre (3,2)(3, -2), radius 55.

PYQ 3 (Practice Test)

Find the distance between (2,3)(2, 3) and (7,15)(7, 15).

d=(72)2+(153)2=25+144=169=13d = \sqrt{(7-2)^2 + (15-3)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

A 5-12-13 Pythagorean triple — SAT loves these.

Difficulty Distribution

Difficulty% of SAT QsTypical type
Easy35%35\%Direct slope, distance plug-ins
Medium50%50\%Circle equations, parallel/perpendicular lines
Hard15%15\%System of equations with geometric interpretation

Expert Strategy

Memorise the Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25. SAT distance questions often use these so the answer is a clean integer.

For circle questions, immediately recognise the standard form. If given general form, complete the square — never plug in points.

For “find the equation of a line through two points” questions, compute slope first, then plug one point into y=mx+by = mx + b.

Common Traps

Using the slope formula upside down: (x2x1)/(y2y1)(x_2 - x_1)/(y_2 - y_1). Slope is rise over run, so yy on top.

Saying perpendicular slopes have product 11. They have product 1-1. Negative reciprocals.

Forgetting to take the square root in distance formula. d2=25+144=169d^2 = 25 + 144 = 169, but d=13d = 13, not 169169.

Treating (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2 centre as (h,k)(-h, -k). The centre is (h,k)(h, k) with the signs as written. So (x3)2(x-3)^2 means h=3h = 3, not 3-3.