Coordinate geometry
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| circle/ˈsɜːkl/ | The set of points at a fixed distance r from its centre: (x − a)² + (y − b)² = r². |
| gradient/ˈɡreɪdɪənt/ | hệ số góc |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | vuông góc |
| midpoint/ˈmɪdpɔɪnt/ | trung điểm |
| perpendicular bisector/ˌpɜːpənˈdɪkjʊlə baɪˈsektə/ | đường trung trực |
| centre/ˈsentə/ | The point equally distant from every point of a circle. |
| radius/ˈreɪdɪəs/ | The distance from the centre of a circle to its edge. |
| diameter/daɪˈæmɪtə/ | A straight line through a circle's centre joining two points on its circumference, or the length of that line. |
| chord/kɔːd/ | A line segment joining two points on a circle. |
| tangent/ˈtændʒənt/ | tiếp tuyến |
Where does a straight path meet a circular boundary?
- On a map, a straight path may cross a circular boundary twice, touch it once, or miss it. Equations let us decide without trusting a rough sketch.
- Start with coordinates and distances. Then connect a line equation, a circle 圆 equation and the number of solutions they share.
Choose a useful line equation
- The gradient 斜率 through two points is $m=\dfrac{y_2-y_1}{x_2-x_1}$ when $x_2\ne x_1$. Use $y=mx+c$, $y-y_1=m(x-x_1)$, or $ax+by+c=0$ for the same line.
- Distinct non-vertical parallel lines have equal gradients. For perpendicular 垂直 lines with finite gradients, $m_1m_2=-1$. A vertical line $x=k$ has undefined gradient; it is perpendicular to a horizontal line $y=h$, whose gradient is zero.
Two points, three forms. Through $(2,3)$ and $(6,11)$, $m=(11-3)/(6-2)=2$.
Substitute both points to check. In a general equation, first rearrange for $y$: $2x-3y+6=0$ has gradient $2/3$, not $2$.
The straight line
y = ax + b
The gradient a tilts the line; the intercept b slides it up and down.
What is the gradient of the line through (2, 3) and (6, 11)?
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m = (11 − 3) / (6 − 2) = 8/4 = 2.
A line has gradient 2. What is the gradient of a line perpendicular to it?
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Perpendicular gradients multiply to −1, so the gradient is −1/2 = −0.5.
The line through P(−3,5) and Q(3,1) is L. Find its gradient and write its equation in point-slope, y=mx+c, and ax+by+c=0 forms. Find both axis intercepts and check both points.
The coefficient of x in an unrearranged general equation is not the gradient.
Through R(4,−2), write the lines (a) parallel to y=7, (b) perpendicular to y=7, (c) parallel to x=−3, and (d) perpendicular to x=−3. Explain their gradients.
A vertical line is written x=constant; it cannot be written y=mx+c with a finite m.
Use distances and midpoints 中点
- Pythagoras gives the distance $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. The midpoint is $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$: average each coordinate separately.
- A perpendicular bisector 垂直平分线 passes through the midpoint at right angles to the segment. Every point on it is equally far from the two ends; the midpoint alone does not give its equation.
For $P(-1,2)$ and $Q(5,6)$, the changes are $6$ and $4$. So $PQ=\sqrt{52}=2\sqrt{13}$ and the midpoint is $M(2,4)$.
The segment gradient is $2/3$, so its perpendicular bisector has gradient $-3/2$:
For a horizontal segment, the perpendicular bisector is vertical: use $x=$ the midpoint's $x$-coordinate instead of taking a reciprocal of zero.
The midpoint of (2, 3) and (8, 7) has x-coordinate:
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Midpoint x = (2 + 8)/2 = 5.
P=(−2,1) and Q=(4,5). Find their distance, midpoint and perpendicular bisector. Show S(3,0) lies on the bisector and check its distances from P and Q.
A perpendicular line through an endpoint is not the perpendicular bisector.
Read and complete a circle equation
- A circle with centre 圆心 $(a,b)$ and positive radius 半径 $r$ has equation $(x-a)^2+(y-b)^2=r^2$. The signs inside the brackets are opposite to the centre coordinates.
- In $x^2+y^2+2gx+2fy+c=0$, completing both squares gives centre $(-g,-f)$ and $r^2=g^2+f^2-c$. A negative value of $r^2$ gives no real circle; zero gives a single point.
Complete both squares, keeping the added constants:
The centre is $(3,-5)$ and the radius is $\sqrt{61}$, not $61$. Substitute any proposed point into the equation to test whether it lies on the circle.
The circle (x − 3)² + (y + 2)² = 25 has what radius?
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r² = 25, so r = 5; the centre is (3, −2).
The circle x² + y² − 6x + 10y − 27 = 0 has centre:
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Complete the square: (x−3)² + (y+5)² = 9 + 25 + 27 = 61. Centre is (3, −5).
Complete the square for x²+y²+4x−6y−12=0. State its centre and radius. Test P(1,7) and Q(2,7). Would replacing the constant −12 by +20 still give a real circle? Explain.
Keep the square-completion constants and take the positive square root for a radius.
Build a circle from a diameter 直径
- The centre is the midpoint of a diameter, and the radius is half its length. For diameter ends $A(0,0)$ and $B(4,2)$, the centre is $C(2,1)$ and $r^2=2^2+1^2=5$.
- Join the diameter ends to any other point on the circle: the angle there is $90^\circ$. Also, the line from the centre to the midpoint of a non-diameter chord 弦 is perpendicular to that chord; this is a line of symmetry.
Here $\angle ADB=90^\circ$ and $CM\perp AD$. The line through $C$ and $M$ reflects $A$ onto $D$.
A chord does not have to pass through the centre. You may use its midpoint as the centre only when the question says the chord is a diameter.
A(−1,2) and B(5,4) are diameter ends. Find the circle equation. Verify D(1,6) lies on it, then state angle ADB with a geometrical reason and confirm using gradients.
An angle subtended by a general chord is not necessarily 90°.
Find a tangent 切线 at a known point
- A tangent is perpendicular to the radius at its point of contact. For the diagram's $B(4,2)$, $m_{CB}=(2-1)/(4-2)=1/2$, so the tangent gradient is $-2$.
- Use the contact point, not the centre, in the tangent equation: $y-2=-2(x-4)$, or $2x+y=10$. This line passes through $B$ but not $C$.
At the top and bottom of $(x-2)^2+(y-1)^2=5$, the radii are vertical, so the tangents are horizontal: $y=1\pm\sqrt5$. At the left and right points, the tangents are vertical: $x=2\pm\sqrt5$. Do not divide by zero.
A tangent to a circle is perpendicular to the radius at the point where it touches.
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This right-angle fact solves most circle problems.
Find unknown points with a given tangent gradient
- Suppose a tangent to $(x-2)^2+(y-1)^2=5$ has gradient $-2$, but its contact point is unknown. Its radius must have gradient $1/2$, so the point lies on the line $y-1=\tfrac12(x-2)$ through the centre.
- Substitute this radius line into the circle. With $u=x-2$, $y-1=u/2$, so $u^2+u^2/4=5$: $u=\pm2$. The two contact points are $(4,2)$ and $(0,0)$, giving tangents $2x+y=10$ and $2x+y=0$.
Given tangent gradient → perpendicular radius line through the centre → substitute into the circle → retain both contact points. Keep exact square roots when the solutions are not integers. This method uses geometry and algebra; no circle differentiation is needed.
Find both points on (x−1)²+(y+2)²=9 where the tangent has gradient −1/2. Give both tangent equations in exact form, using radius geometry and algebra.
The gradient fixes a radius direction, not a unique contact point. The two opposite points give parallel tangents.
Count line–circle intersections
- Substitute the line into the circle and solve for one coordinate, then recover the other. For $y=0$ in $(x-2)^2+(y-1)^2=5$, $(x-2)^2=4$, so the intersections are $(0,0)$ and $(4,0)$.
- For a variable line $y=k$, the same circle gives $(x-2)^2=5-(k-1)^2$. A positive right side gives two points; zero gives one tangent point; a negative right side gives none.
The boundaries are $k=1-\sqrt5$ and $k=1+\sqrt5$.
- Two intersections when $1-\sqrt5<k<1+\sqrt5$.
- One intersection at either boundary, with $x=2$.
- No intersections outside that interval.
For a sloping line, substitution gives a quadratic too. Its discriminant gives the same two/one/none test. A repeated root represents one point, not two different points.
For the circle (x+1)²+(y−2)²=9 and line y=k, find all k giving two, one, or no intersections. Give the tangent points at the boundaries and the intersections when k=2.
The parameter is a y-coordinate here; the tangent boundaries are centre y plus or minus radius.
Connect line–curve graphs to algebra
- Intersections share both coordinates, so equate the two expressions for $y$. For $y=x^2$ and $y=2x+k$, solve $x^2-2x-k=0$; its discriminant is $4+4k$.
- There are two intersections for $k>-1$, one for $k=-1$ and none for $k<-1$. At the boundary, $x=1$ gives the single point $(1,1)$. Always classify the discriminant you actually obtain: some line families intersect a curve for every real parameter value.
A sketch suggests what may happen; it does not prove an exact parameter boundary. Show the equation, the discriminant condition and its solution. If the coefficient of $x^2$ can vanish, check that case separately before using the quadratic test.
Show that y=2x²−3x+4 and y=kx+6 intersect at two distinct points for every real k. Explain why there is no interval of k with no intersection.
Do not force a parameter interval: classify the expression the actual problem gives.
Turn distances from axes into a centre
- A circle entirely in the first quadrant has centre $(a,b)$ and radius $r$. Its least distance from the $y$-axis is $a-r$; from the $x$-axis it is $b-r$. The axis names refer to distances perpendicular to the axes.
- If those gaps are $1$ and $3$, then the centre is $(r+1,r+3)$. If its distance from the origin is $\sqrt{34}$, Pythagoras gives the radius.
Reject $r=-6$: a radius is positive. Thus $r=2$ and the centre is $(3,5)$. The point furthest from the origin lies along the ray from the origin through the centre. Its radius has gradient $5/3$, so the tangent there has gradient $-3/5$.
A circle lies entirely in the first quadrant. Its least distance from the x-axis is 6 and from the y-axis is 2. The distance from the origin to its centre is 4√5. Find its radius, centre and equation. Find the tangent gradient at its point furthest from the origin.
A distance to the x-axis constrains the y-coordinate. The nearest point to an axis is one radius closer than the centre.
Choose and check a complete method
- Use a line form suited to the information; handle horizontal and vertical lines directly.
- Complete both squares for a circle, or use a diameter's midpoint and half-length.
- Use radius–tangent perpendicularity, semicircle angles and symmetry where they apply.
- Substitute for intersections; count real solutions and state exact parameter boundaries.
Write an independent solution
- For each task, show equations and coordinate pairs. Check a line with a known point and a circle with a known point on its circumference.
- For a parameter condition or a rejected value, explain the boundary or the reason. A correct number without the method is not a complete solution.
Find the gradient and equation of the line through (0, 1) and (3, 10). What is the radius of (x - 1) squared plus (y + 2) squared = 9?
9 divided by 3 is 3. The square root of 9 is 3.
A line passes through (0,1) and (3,10). What is its gradient?
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10 minus 1, divided by 3, is 3.